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5th Year Applied Maths Higher Level Kieran Mills
Uniform Accelerated Motion
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Ref: 5/appmaths/h/km/UAM
Oral Preparation CoursesSeparate to the Easter Revision Courses, The Dublin School of Grinds is also running Oral Preparation Courses. With the Oral marking component of the Leaving Certificate worth up to 40%, it is of paramount importance that students are fully prepared for these examinations. These courses will show students how to lead the Examiner towards topics that the student is prepared in. This will provide students with the confidence they need to perform at their peak.
ORAL PREPARATION COURSE FEES:
PRICE TOTAL SAVINGS
1st Oral Course €140 €140 -
2nd Oral Course €100 €240 €40
Looking to maximise your CAO points?Easter is well known as a time for students to vastly improve on the points that they received in their mock exams. To help students take advantage of this valuable time, The Dublin School of Grinds is running intensive exam-focused Easter Revision Courses. Each course runs for five days (90 minutes per day).
The focus of these courses is to maximise students’ CAO points. Special offer: Buy 1st course and get 2nd course free. To avail of this offer, early booking is required as courses were fully booked last year.
What do students get at these courses?
9 90 minutes of intensive tuition per day for five days, with Ireland’s leading teachers.
9 Comprehensive study notes.
9 A focus on simple shortcuts to raise students’ grades and exploit the critically important marking scheme.
9 Access to a free supervised study room.
9 Access to food and beverage facilities.
EASTERREVISION COURSES EASTER REVISION COURSE FEES:
PRICE TOTAL SAVINGS
1st Course €295 €295 -
2nd Course FREE €295 €295
3rd Course €100 €395 €490
4th Course €100 €495 €685
5th Course €100 €595 €880
6th Course €100 €695 €1,075
7th Course €100 €795 €1,270
8th Course €100 €895 €1,465
9th Course €100 €995 €1,660
To book, call us on 01-442 4442 or book online at www.dublinschoolofgrinds.ie
NOTE: These courses are built on the fact that there are certain predicable trends that appear and reoccur over and over again in the State Examinations.
FREE DAILY BUS SERVICE For full information on our Easter bus service, see 3 pages ahead.
NOTE: Any bookings for Junior Cert courses will also receive a weekly grind in one subject for the rest of the academic year, free of charge. This offer applies to 3rd and 2nd year students ONLY.
Timetable An extensive range of course options are available over a two-week period to cater for students’ timetable needs. Courses are held over the following weeks:
» Monday 21st March – Friday 25th March 2016» Monday 28th March – Friday 1st April 2016
All Easter Revision Courses take place in The Talbot Hotel, Stillorgan (formerly known as The Stillorgan Park Hotel).
BOOK EARLY TO AVAIL OF THE SPECIAL OFFER
BUY 1ST COURSE GET 2ND COURSE
F R E E ! Due to large course content, these subjects have been
divided into two courses. For a full list of topics covered in these courses, please see 3 pages ahead.
*
6th Year Easter Revision CoursesSUBJECT LEVEL DATES TIME
Accounting H Monday 21st March – Friday 25th March 8:00am - 9:30am
Agricultural Science H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
Applied Maths H Monday 28th March – Friday 1st April 8:00am - 9:30am
Art History H Monday 28th March – Friday 1 April 8:00am - 9:30am
Biology Course A* H Monday 21st March – Friday 25th March 8:00am - 9:30am
Biology Course A* H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
Biology Course A* H Monday 28th March – Friday 1st April 10:00am - 11:30am
Biology Course B* H Monday 21st March – Friday 25th March 10:00am - 11:30am
Biology Course B* H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
Biology Course B* H Monday 28th March – Friday 1st April 8:00am - 9:30am
Business H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
Business H Monday 28th March – Friday 1st April 8:00am - 9:30am
Chemistry Course A* H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Chemistry Course B* H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
Classical Studies H Monday 21st March – Friday 25th March 8:00am - 9:30am
Economics H Monday 21st March – Friday 25th March 8:00am - 9:30am
Economics H Monday 28th March – Friday 1st April 10:00am - 11:30am
English Paper 1* H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
English Paper 2* H Monday 21st March – Friday 25th March 10:00am - 11:30am
English Paper 2* H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
English Paper 2* H Monday 28th March – Friday 1st April 10:00am - 11:30am
English Paper 2* H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
French H Monday 21st March – Friday 25th March 10:00am - 11:30am
French H Monday 28th March – Friday 1st April 8:00am - 9:30am
Geography H Monday 28th March – Friday 1st April 8:00am - 9:30am
Geography H Monday 28th March – Friday 1st April 10:00am - 11:30am
German H Monday 21st March – Friday 25th March 10:00am - 11:30am
History (Europe)* H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
History (Ireland)* H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
Home Economics H Monday 21st March – Friday 25th March 10:00am - 11:30am
Irish H Monday 21st March – Friday 25th March 10:00am - 11:30am
Irish H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Maths Paper 1* H Monday 21st March – Friday 25th March 8:00am - 9:30am
Maths Paper 1* H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
Maths Paper 1* H Monday 28th March – Friday 1st April 10:00am - 11:30am
Maths Paper 1* H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
Maths Paper 2* H Monday 21st March – Friday 25th March 10:00am - 11:30am
Maths Paper 2* H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
Maths Paper 2* H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Maths Paper 2* H Monday 28th March – Friday 1st April 4:00pm - 5:30pm
Maths O Monday 21st March – Friday 25th March 8:00am - 9:30am
Maths O Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Physics H Monday 28th March – Friday 1st April 10:00am - 11:30am
Spanish H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
Spanish H Monday 28th March – Friday 1st April 10:00am - 11:30am
6th Year Oral Preparation CoursesSUBJECT LEVEL DATES TIME
French H Sunday 20th March 10:00am - 2:00pm
German H Saturday 26th March 10:00am - 2:00pm
Irish H Saturday 26th March 10:00am - 2:00pm
Spanish H Saturday 19th March 1:00pm - 5:00pm
5th Year Easter Revision CoursesSUBJECT LEVEL DATES TIME
Maths H Monday 28th March – Friday 1st April 8:00am - 9:30am
English H Monday 28th March – Friday 1st April 4:00pm - 5:30pm
Note: 5th year students are welcome to attend any 6th year course as part of our buy 1 get 1 free offer.
3rd Year Easter Revision CoursesSUBJECT LEVEL DATES TIME
Business Studies H Monday 28th March – Friday 1st April 8:00am - 9:30am
English H Monday 21st March – Friday 25th March 8:00am - 9:30am
English H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
French H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Geography H Monday 28th March – Friday 1st April 12:00pm - 1:30pm
German H Monday 21st March – Friday 25th March 8:00am - 9:30am
History H Monday 21st March – Friday 25th March 4:00pm - 5:30pm
Irish H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
Maths H Monday 21st March – Friday 25th March 10:00am - 11:30am
Maths H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
Maths H Monday 28th March – Friday 1st April 10:00am - 11:30am
Maths O Monday 28th March – Friday 1st April 12:00pm - 1:30pm
Science H Monday 28th March – Friday 1st April 2:00pm - 3:30pm
Science H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
Spanish H Monday 21st March – Friday 25th March 12:00pm - 1:30pm
2nd Year Easter Revision CoursesSUBJECT LEVEL DATES TIME
Maths H Monday 21st March – Friday 25th March 2:00pm - 3:30pm
NOTE: Any bookings for Junior Cert courses will also receive a weekly grind in one subject for the rest of the academic year, free of charge. This offer applies to 3rd and 2nd year students ONLY.
Contents: Uniform ACCelerAted motion
seCtion 1 Using the FormUlae ...................................................................2Exercise 1 ..................................................................................6Exercise 2 ..................................................................................8
seCtion 2 sUccessive times and distances ................................................10Exercise 3 ..................................................................................14
seCtion 3 catch-Up problems ....................................................................16Exercise 4 ..................................................................................25
seCtion 4 velocity-time graphs ................................................................28Exercise 5 ..................................................................................39
seCtion 5 Free-Fall ....................................................................................42Exercise 6 ..................................................................................46
leaving cert QUestions 2014-1996 .............................................................49
©The Dublin School of Grinds Page 1 Kieran Mills & Tony Kelly
Sec
tio
n 1
: USi
ng
th
e F
or
mU
la
e
Qu
an
tity
Sy
mb
ol
un
itS
Acc
eler
atio
n a
m s– 2
Initi
al v
eloc
ity
u m
s– 1
Fina
l vel
ocity
v
m s– 1
Tim
e t
sD
ispl
acem
ent
s m
eq
Uat
ion
S oF
mo
tio
n
Ex.
How
long
doe
s it t
ake
the
bus t
o tra
vel f
rom
A to
B?
How
far i
s it f
rom
A to
B?
Sol
Ut
ion
a u v t
= = = =
−
− −
4 10 50
2 1 1
ms ms
ms
?
vuatt
tt=
+=
+=
∴=
5010
440
4 10s
svut
=+
=+
= =
1 2 1 2 1 2
5010
1060
1030
0()
()(
)(
)()
m
vuat
ss
vut
asut
at
=+
=+
=+
......
...(
)[]
().
...(
)[]
...
1 2N
oN
o1 2
1 22
..()[
]
...(
)[]
....(
)[]
3 4 5
No
No
No
vv
uas
tsvt
a tu
22
1 22
2=
+
=−
0 m
v =
50
m s
-1
AB
u =
10
m s
-1
s =
30
0 m
t =
10
s
a =
4 m
s-2
-R
ight
+Le
ft
+a:
acce
lera
tion
–a:
dece
lera
tion
©The Dublin School of Grinds Page 2 Kieran Mills & Tony Kelly
ex
am
ple 1
u =
3 m
s–1, a
= 6
m s–2
, t =
4 s,
s =
?
ex
am
ple 2
u =
6 m
s–1, a
= 1
0 m
s–2, t
= 4
s, v
= ?
ex
am
ple 3
u =
8 m
s–1, v
= 1
5 m
s–1, t
= 6
s, s
= ?
ex
am
ple 4
u =
4 m
s–1, v
= 9
m s–1
, a =
0.5
m s–2
, s =
?
ex
am
ple 5
s = 1
20 m
, v =
10
m s–1
, t =
3 s,
a =
?
ex
am
ple 6
s = 8
m, u
= 6
m s–1
, a =
–2
m s–2
, t =
?
©The Dublin School of Grinds Page 3 Kieran Mills & Tony Kelly
ex
am
ple 1
u =
3 m
s–1, a
= 6
m s–2
, t =
4 s,
s =
?So
lU
tio
n
mat
he
mat
ica
l c
al
cU
lat
ion
S
u =
3 m
s–1
a =
6 m
s–2
t = 4
ss =
?
No
v pr
esen
t. U
se E
quat
ion
(3).
sut
at=
+
=+
=+
=
1 22 1 2
23
46
412
4860(
)()
()(
)
m
ex
am
ple 2
u =
6 m
s–1, a
= 1
0 m
s–2, t
= 4
s, v
= ?
Sol
Ut
ion
mat
he
mat
ica
l c
al
cU
lat
ion
S
u =
6 m
s–1
a =
10 m
s–2
t = 4
sv
= ?
No
s pre
sent
. Use
Equ
atio
n (1
).
ex
am
ple 3
u =
8 m
s–1, v
= 1
5 m
s–1, t
= 6
s, s
= ?
Sol
Ut
ion
mat
he
mat
ica
l c
al
cU
lat
ion
S
u =
8 m
s–1
v =
15 m
s–1
t = 6
ss =
?
No
a pr
esen
t. U
se E
quat
ion
(2).
vuat
=+
=+
=+
=−
610
46
4046
1()
ms
svut
=+
=+
= =
1 2 1 215
86
323
69()
()
() m
©The Dublin School of Grinds Page 4 Kieran Mills & Tony Kelly
ex
am
ple 4
u =
4 m
s–1, v
= 9
m s–1
, a
= 0.
5 m
s–2, s
= ?
Sol
Ut
ion
mat
he
mat
ica
l c
al
cU
lat
ion
S
u =
4 m
s–1
v =
9 m
s–1
a =
0.5
m s–2
s = ?
No
t pre
sent
. Use
Equ
atio
n (4
).
ex
am
ple 5
s = 1
20 m
, v =
10
m s–1
, t =
3 s,
a =
?So
lU
tio
n
mat
he
mat
ica
l c
al
cU
lat
ion
S
s = 1
20 m
v =
10 m
s–2
t = 3
sa
= ?
No
u pr
esen
t. U
se E
quat
ion
(5).
ex
am
ple 6
s = 8
m, u
= 6
m s–1
, a =
–2
m s–2
, t =
?So
lU
tio
n
mat
he
mat
ica
l c
al
cU
lat
ion
S
s = 8
mu
= 6
m s–1
a =
–2 m
s–2
t = ?
No
v pr
esen
t. U
se E
quat
ion
(3).
vu
ass
ss2
2
22
29
42
05
8116 65=
+
=+
=+
=
(.
)
m
svt
ata
a aa a=
−
=−
=−
=−
=−
=−
1 22
1 22
9 2
120
103
312
030
240
609
918
020
()
()
mms−
2
sut
att
ttt
tt
tt
t
=+
=+
−
=−
−+
=−
−=
=
1 22
1 22
2
286
2
86 6
80
42
02(
)(
)
()(
),s
44s
©The Dublin School of Grinds Page 5 Kieran Mills & Tony Kelly
1. u a t s= = = =− −0 4 61 2m s m s s, , , ?
2. u a t v= = = =− −5 3 31 2m s m s s, , , ?
3. u a t s= = = =− −0 4 41 2m s m s s, , , ?
4. u v t s= = = =− −7 15 101 1m s m s s, , , ?
5. u v s a= = = =− −4 10 71 1m s m s m, , , ?
6. v a t u= = = =− −30 6 21 2m s m s s, , , ?
7. u v s a= = = =− −6 3 1 51 1m s m s m, , . , ?
8. s a t u= = = =−35 6 22m, m s s, , ?
9. s u v t= = = =− −40 5 151 1m m s m s, , , ?
10. u v a s= = = =− − −5 10 0 51 1 2m s m s m s, , . , ?
11. u v t a= = = =− −48 12 91 1m s m s s, , , ?
12. s t a v= = = =−30 2 4 2m s m s, , , ?
13. s t v a= = = =−120 3 10 1m s m s, , , ?
14. s a v t= − = = =− −4 2 32 1m m s m s, , , ?
15. s v a t= = = =− −8 7 31 2m m s m s, , , ?
16. s u a t= = = =− −24 5 21 2m m s m s, , , ?
17. s u a t= = = − =− −8 6 21 2m m s m s, , , ?
18. v a t u= = = =− −72 0 5 11 2km h m s, . , min, ?
19. s v a u= = = =− −8 8 31 2m m s m s, , , ?
20. s v a t= − = = =− −6 4 41 2m m s m s, , , ?
ExErcisE 1. Using thE formUlaE
©The Dublin School of Grinds Page 6 Kieran Mills & Tony Kelly
answErs
ExErcisE 11. 72 m
2. 14 m s–1
3. 32 m
4. 110 m
5. 6 m s–2
6. 18 m s–1
7. –9 m s–2
8. 11.5 m s–1
9. 4 s
10. 75 m
11. –4 m s–2
12. 19 m s–1
13. –20 m s–2
14. 4 s
15. 2 83s s,
16. 3 s
17. 2s, 4s
18. –10 m s–1
19. 4 m s–1
20. 3 s
©The Dublin School of Grinds Page 7 Kieran Mills & Tony Kelly
ExErcisE 2. simplE problEms Using thE formUlaE
1. A train starts from rest and accelerates uniformly at 2.5 m s–2 until it reaches a speed of25 m s–1. Find the distance moved and the time taken for this motion.
2. A car can accelerate from rest to 90 km h–1 in 7.5 seconds. Find its acceleration.
3. In travelling 65 cm along the barrel of a rifle a bullet accelerates from rest to 230 m s–1.Find the accceleration and the time the bullet is in the barrel.
4. A car travelling at 24 m s–1 requires a minimum braking distance of 36 m. What is itsdeceleration? How long does it take to stop?
5. A car starts from rest with acceleration 4 m s–2. How far does it go in (i) 2 s, (ii) 3 s,(iii) the third second.
6. A body moves in a straight line and increases its velocity from 3 m s–1 to 15 m s–1
uniformly in 6 s. Find the acceleration and the distance travelled.
7. A particle starts with a velocity of 3 m s–1 and accelerates uniformly at 1.5 m s–2.How far does it go in (i) 1 s, (ii) 5 s, (iii) the fourth second.
8. A body is projected from the origin with a velocity of 8 m s–1 and acceleration –2 m s–2.Find(i) the velocity when t = 3 s,(ii) when it comes to instantaneous rest.
9. A particle moves along a straight line between two points P and Q with constantacceleration 0.8 m s–2. Its velocity at Q is 1.2 m s–1 greater than the velocity at P.If the distance PQ is 48 m, find the velocity at P.How long after passing P does it take the velocity to reach 48 m s–1.
10. A car is moving with speed u m s–1. The brakes of the car can produce a constantdeceleration of 5 m s–2. It is known that when the driver decides to stop, a period of 2
5 selapses before the brakes are applied. As the car passes a point O, the driver decides tostop.Find in terms of u the minimum distance of the car from O when the car comes to rest.The driver is approaching traffic lights and is 102 m away when the light changes fromgreen to amber. The lights remain amber for 3 s before changing to red.Show(a) when u < 30 the driver can stop before reaching the lights,(b) when u > 34 the driver can pass the light before it turns red.
©The Dublin School of Grinds Page 8 Kieran Mills & Tony Kelly
answErs
ExErcisE 21. 125 m, 10 s
2. 103
2ms−
3. 40, 692 m s–1, 5 652 10 3. × − s
4. –8 m s–2, 3 s
5. 8 m, 18 m, 10 m
6. 2 m s–2, 54 m
7. 3.75 m, 33.75 m, 8.25 m
8. 2 m s–1, 4 s
9. 31.4 m s–1, 20.75 s
10. 110
2 25u u+
©The Dublin School of Grinds Page 9 Kieran Mills & Tony Kelly
Sec
tio
n 2
: Su
cc
eSS
ive t
ime
S an
d d
iSta
nc
eS
ex
am
ple 1
In tw
o su
cces
sive
seco
nds a
uni
form
ly a
ccel
erat
ing
body
trav
els 5
m a
nd
13 m
. Fin
d its
acc
eler
atio
n an
d its
initi
al v
eloc
ity.
ex
am
ple 2
A u
nifo
rmly
dec
eler
atin
g bo
dy c
over
s suc
cess
ive
100
m d
ista
nces
in 5
s an
d 10
s. F
ind
its in
itial
spee
d, th
e de
cele
ratio
n an
d th
e fu
rther
tim
e fo
r th
e bo
dy to
com
e to
rest
.
©The Dublin School of Grinds Page 10 Kieran Mills & Tony Kelly
ex
am
ple 1
In tw
o su
cces
sive
seco
nds a
uni
form
ly a
ccel
erat
ing
body
trav
els 5
m a
nd 1
3 m
. Fin
d its
acc
eler
atio
n an
d its
initi
al
velo
city
.So
lu
tio
n
a
t =
1 s
t =
2 s
, s =
18
m
t =
1 s
us =
5 m
s =
13
m
sut
at=
+1 2
2
182
218
22
1 22
=+
=+
ua
ua
()
()
....(
)2
51
15 10
2
1 22
1 2
=+
=+
=+
ua
ua
ua
()
()
....(
)1
In th
ese
type
s of p
robl
ems a
lway
s use
the
equa
tion
show
n.
Take
all
valu
es fr
om th
e be
ginn
ing
poin
t.m
ath
em
atic
al c
al
cu
lat
ion
S
An
swer
s
102
118
22
102
182
28
10
=+
×−
=+
−=−
−=
+=
ua
ua
ua
ua
a
......
()(
)...
.()1 2
==+
=+
=∴
=
==
−−
210
28
22
1
81
21
ua
u uu
au....(
)
,
1
ms
ms
©The Dublin School of Grinds Page 11 Kieran Mills & Tony Kelly
sut
at=
+1 2
2
In th
ese
type
s of p
robl
ems a
lway
s use
the
equa
tion
show
n.
Take
all
valu
es fr
om th
e be
ginn
ing
poin
t.
ut =
5 s
t =
15
s, s
= 2
00
m
s =
10
0 m
s =
10
0 m
t =
10
st =
?
a
0 m
s-1
100
55
100
540
25
1 22
25 2
=+
=+
=+
ua
ua
ua
()
()
.....(
)1
200
1515
200
1580
645
1 22
225 2
=+
=+
=+u
au
au
a
()
()
.....(
)2
mat
he
mat
ica
l c
al
cu
lat
ion
S
402
53
806
45
120
615
806
=+
×−
=+
−=−
−=
+
ua
ua
ua
u
......
.()(
)...
..()1 2
44540
30
402
540
25
402
120
62
4 3
20 3
aa
ua
u u u
−=
=+
=+
−
=−
=−
......
.()
()1
0014
06
140 6
70 31
= =
∴=
−
uu
um
s
co
nt...
.
ex
am
ple 2
A u
nifo
rmly
dec
eler
atin
g bo
dy c
over
s suc
cess
ive
100
m d
ista
nces
in 5
s an
d 10
s. F
ind
its in
itial
spee
d, th
e de
cele
ratio
n an
d th
e fu
rther
tim
e fo
r the
bod
y to
com
e to
rest
.So
lu
tio
n
©The Dublin School of Grinds Page 12 Kieran Mills & Tony Kelly
ex
am
ple 2
A u
nifo
rmly
dec
eler
atin
g bo
dy c
over
s suc
cess
ive
100
m d
ista
nces
in 5
s an
d 10
s. F
ind
its in
itial
spee
d, th
e de
cele
ratio
n an
d th
e fu
rther
tim
e fo
r the
bod
y to
com
e to
rest
.So
lu
tio
n
sut
at=
+1 2
2
ut =
5 s
t =
15
s, s
= 2
00
m
s =
10
0 m
s =
10
0 m
t =
10
st =
?
a
0 m
s-1
mat
he
mat
ica
l c
al
cu
lat
ion
S
402
53
806
45
120
615
806
=+
×−
=+
−=−
−=
+
ua
ua
ua
u
......
.()(
)...
..()1 2
44540
30
402
540
25
402
120
62
4 3
20 3
aa
ua
u u u
−=
=+
=+
−
=−
=−
......
.()
()1
0014
06
140 6
70 31
= =
∴=
−
uu
um
s
mat
he
mat
ica
l c
al
cu
lat
ion
S
An
swer
: Fur
ther
tim
e =
17.5
– 1
5 =
2.5
s
Furth
er ti
me
for b
ody
to c
ome
to re
st:
u a v t
= =−
= =
−
−
−
70 31
4 32
10
ms ms
ms
?
vuat
tt t t
=+
=+−
= = ==
0 470
175
70 34 3
4 370 3
70 4
()
.s
©The Dublin School of Grinds Page 13 Kieran Mills & Tony Kelly
ExErcisE 3. succEssivE TimEs/succEssivE DisTancEs
1. In two successive seconds a uniformly accelerating body travels 4 m and 8 m.Find its acceleration.
2. A uniformly accelerating body travels 5 m and 11 m repectively in its first two seconds.How far does it travel in the fourth second?
3. A uniformly decelerating body covers successive 100 m distances in 5 s and 10 s.Find its initial speed, the deceleration and the further time for the body to come to rest.
4. A particle starts from rest and moves in a straight line with uniform acceleration.It passes three points A, B and C where |AB| =105 m and |BC| = 63 m. If it takes 6 s to travelfrom A to B and 2 s from B to C find(i) its acceleration, (ii) the distance of A from the starting position.
5. A sprinter runs a race with constant acceleration throughout. During the race he passes fourposts A, B, C, D such that |AB| = |BC| = |CD| = 36 m. If the sprinter takes 3 s to run from A toB and 2 s to run from B to C, how long does it takes to run from C to D?
6. A particle moving in a straight line with uniform acceleration describes 23 m in the fifthsecond of its motion and 31 m in the seventh second. Calculate its initial velocity.
7. A body travels in a straight line with uniform acceleration. The particle passes three points A,B and C at t = 0, t = 3 s and t = 6 s. If |BC| = 90 m and the speed of the particle at B is21 m s–1, find the acceleration of the body and its speed at A.
8. A, B, C are three points which lie in that order on a straight road with |AB| = 45 m and|BC| = 32 m. A car travels along the road in the direction ABC with constant acceleration f.The car passes A with speed u and passes B five seconds later and passes C two seconds afterthat. Find u and f.
9. A car is moving along at a steady 20 m s–1 when the driver suddenly sees a tree across theroad 56 m ahead. He immediately applies the brakes giving the car a constant deceleration of4 m s–2. How far in front of the tree does the car come to rest? If the driver had not reactedimmediately and the brakes were applied one second later with what speed would the carhave hit the tree?
10. A, B, C are three points on a straight line in that order. A body is projected from B towards Awith a speed of 5 m s–1. The body experiences an acceleration of 2 m s–2 towards C.If |BC| = 24 m, find the time to reach C, and the distance travelled by the body from theinstant of projection until it reaches C.
11. A bus 12.5 m long travels with constant acceleration. The front of the bus passes a point Pwith speed u and the rear passes P with speed v. Find in terms of u and v(i) the time taken for the bus to pass P,(ii) what fraction of the bus passes P in half this time.
©The Dublin School of Grinds Page 14 Kieran Mills & Tony Kelly
12. A body moving in a straight line with constant acceleration passes in succession throughpoints A, B, C and D where |AB| = x, |BC| = y and |CD| = z where the distances x, y and z arecovered in equal intervals of time. Show 2y = x + z.
13. A uniformly decelerating train of length 40 m enters a station of length 80 m. The frontengine leaves the station 5 s later and the rear of the train leaves the station after a further5 s. Find the deceleration of the train.
14. A uniformly accelerating body starts with a speed of u, in successive times of t travels
distances s and 2s. Prove that its acceleration is 4 2us
.
15. A body starts moving in a straight line with velocity u and acceleration a. If when thevelocity has increased to 5u the acceleration is reversed in direction its magnitude beingunaltered prove that when the particle returns to its starting point its velocity will be –7u.
answErs
ExErcisE 31. 4 m s–2
2. 23 m
3. 703
1 43
2 2 5ms ms s− −−, , .
4. (i) 3.5 m s–2, (ii) 7 m
5. 1.6 s
6. 5 m s–1
7. 6 m s–2, 3 m s–1
8. 4 m s–1, 2 m s–2
9. 6 m, 10.6 m s–1
10. 8 s, 36.5 m
11. 25 3
4v uu vu v+++
,( )
13. –1.6 m s–2
©The Dublin School of Grinds Page 15 Kieran Mills & Tony Kelly
Sec
tio
n 3
: cat
ch
-up
pro
bl
em
S
ex
am
ple 1
Two
bodi
es st
art t
oget
her a
t the
sam
e tim
e at
the
sam
e pl
ace
and
mov
e al
ong
the
sam
e st
raig
ht li
ne. I
f one
mov
es w
ith a
con
stan
t spe
ed o
f 16
m s–1
whi
le th
e ot
her s
tarts
from
rest
and
mov
es a
t a c
onst
ant
acce
lera
tion
of 4
m s–2
. How
long
will
it ta
ke b
efor
e th
ey a
re to
geth
er?
ex
am
ple 2
Two
bodi
es A
and
B tr
avel
in th
e sa
me
dire
ctio
n al
ong
the
sam
e lin
e.
Bod
y A
star
ts w
ith v
eloc
ity 5
m s–1
and
acc
eler
atio
n 3
m s–2
. The
oth
er
body
star
ts fr
om th
e sa
me
plac
e w
ith v
eloc
ity 2
m s–1
and
acc
eler
atio
n 4
m s–2
. Fin
d w
hen
and
whe
re th
ey a
re to
geth
er a
gain
.
ex
am
ple 3
Two
bodi
es m
ove
in th
e sa
me
dire
ctio
n al
ong
para
llel p
aths
. A st
arts
fr
om a
poi
nt O
with
vel
ocity
8 m
s–1 a
nd a
ccel
erat
ion
2 m
s–2 a
nd B
star
ts
8 m
ahe
ad o
f A a
nd m
oves
off
with
vel
ocity
2 m
s–1 a
t acc
eler
atio
n
4 m
s–2. F
ind
whe
n th
ey w
ill b
e to
geth
er a
nd th
eir d
ista
nces
from
O a
t th
ese
times
. Wha
t are
thei
r res
pect
ive
spee
ds w
hen
they
are
toge
ther
?
A
B
64m
u =
10 m
s–1
a =
2 m
s–2
u =
1 m
s–1
a =
4 m
s–2
ex
am
ple 4
If A
star
ts 2
seco
nds b
efor
e B
find
whe
n an
d w
here
they
are
toge
ther
. Fi
nd th
e m
axim
um d
ista
nce
that
A m
oves
ahe
ad o
f B in
the
subs
eque
nt
mot
ion.
©The Dublin School of Grinds Page 16 Kieran Mills & Tony Kelly
ex
am
ple 1
Two
bodi
es st
art t
oget
her a
t the
sam
e tim
e at
the
sam
e pl
ace
and
mov
e al
ong
the
sam
e st
raig
ht li
ne. I
f one
mov
es
with
a c
onst
ant s
peed
of 1
6 m
s–1 w
hile
the
othe
r sta
rts fr
om re
st a
nd m
oves
at a
con
stan
t acc
eler
atio
n of
4 m
s–2.
How
long
will
it ta
ke b
efor
e th
ey a
re to
geth
er?
Sol
ut
ion
mat
he
mat
ica
l c
al
cu
lat
ion
S
sut
at=
+1 2
2
An
swer
s
u =
16
m s
a =
0 m
s
-1 -2
s=
sA
B
A B
u =
0 m
sa
= 4
m s
-1 -2
tt
t =
0 s
st
tt
st
tt
A B
=+
= =+
=
160
16 04
2
1 22
1 22
2
()
()
()
ss
tt
tt
tt
tttA
B−
=
−=
−=
−=
−=
=
0
162
02
160
80
80
082
2
2 () ,s
s
sepA
rAt
ion
eq
uAt
ion
: s
sA
B−
tog
eth
er (l
evel
): s
sA
B−
=0
A a
nd B
are
toge
ther
afte
r 0 s
and
afte
r 8 s.
ss
tt
AB
−=
−16
22
©The Dublin School of Grinds Page 17 Kieran Mills & Tony Kelly
ex
am
ple 2
Two
bodi
es A
and
B tr
avel
in th
e sa
me
dire
ctio
n al
ong
the
sam
e lin
e. B
ody
A st
arts
with
vel
ocity
5 m
s–1 a
nd
acce
lera
tion
3 m
s–2. T
he o
ther
bod
y st
arts
from
the
sam
e pl
ace
with
vel
ocity
2 m
s–1 a
nd a
ccel
erat
ion
4 m
s–2. F
ind
whe
n an
d w
here
they
are
toge
ther
aga
in.
Sol
ut
ion
mat
he
mat
ica
l c
al
cu
lat
ion
S
sut
at=
+1 2
2
u =
5 m
sa
= 3
m s
-1 -2
s=
sA
B
A B
u =
2 m
sa
= 4
m s
-1 -2
tt
t =
0 s
st
tt
t
st
tt
t
A B
=+
=+
=+
=+
()
()
()
()
53
5 24
22
1 22
3 22 1 2
2
2
30
60
60
06
1 22
2tt
tt
tt
t
−=
−=
−=
=() ,s
s
sepA
rAt
ion
eq
uAt
ion
: s
sA
B−
tog
eth
er (l
evel
): s
sA
B−
=0
ss
tt
tt
tt
AB
−=
+−
−
=−
52
2
3
3 22
2
1 22
A a
nd B
are
toge
ther
afte
r 0 s
and
afte
r 6 s.
Dis
tanc
e A a
nd B
hav
e tra
velle
d af
ter 6
s:s
tt
B=
+
=+
=+
=+
=
22
26
26
26
236
1272
84
2
2(
)(
)(
)(
)
m
©The Dublin School of Grinds Page 18 Kieran Mills & Tony Kelly
ex
am
ple 3
Two
bodi
es m
ove
in th
e sa
me
dire
ctio
n al
ong
para
llel p
aths
. A st
arts
from
a p
oint
O w
ith v
eloc
ity 8
m s–1
and
ac
cele
ratio
n 2
m s–2
and
B st
arts
8 m
ahe
ad o
f A a
nd m
oves
off
with
vel
ocity
2 m
s–1 a
t acc
eler
atio
n 4
m s–2
. Fin
d w
hen
they
will
be
toge
ther
and
thei
r dis
tanc
es fr
om O
at t
hese
tim
es. W
hat a
re th
eir r
espe
ctiv
e sp
eeds
whe
n th
ey
are
toge
ther
?So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
Ssut
at=
+1 2
2B
star
ts 8
m a
head
of A
. The
refo
re, a
dd
8 to
the
dist
ance
equ
atio
n fo
r B.
u =
8 m
sa
= 2
m s
-1 -2
s=
sA
B
A
B u =
2 m
sa
= 4
m s
-1 -2
o
8 m
tt
t =
0 s
st
ttt
st
tt
t
A B
=+
=+
=+
+
=+
+
()
()
()
()
82
8 82
4
82
2
1 22
2
1 22
2
−+
−=
−+
=−
−=
=tt
tt
tt
t
2
2
68
06
80
24
02
4(
)()
,ss
Co
nt.
....
sepA
rAt
ion
eq
uAt
ion
: s
sA
B−
tog
eth
er (l
evel
): s
sA
B−
=0
ss
tt
tt
tt
AB
−=
+−
−−
=−
+−
88
22
68
22
2
©The Dublin School of Grinds Page 19 Kieran Mills & Tony Kelly
ex
am
ple 3
Two
bodi
es m
ove
in th
e sa
me
dire
ctio
n al
ong
para
llel p
aths
. A st
arts
from
a p
oint
O w
ith v
eloc
ity 8
m s–1
and
ac
cele
ratio
n 2
m s–2
and
B st
arts
8 m
ahe
ad o
f A a
nd m
oves
off
with
vel
ocity
2 m
s–1 a
t acc
eler
atio
n 4
m s–2
. Fin
d w
hen
they
will
be
toge
ther
and
thei
r dis
tanc
es fr
om O
at t
hese
tim
es. W
hat a
re th
eir r
espe
ctiv
e sp
eeds
whe
n th
ey
are
toge
ther
?So
lu
tio
n
sut
at=
+1 2
2vuat
=+
u =
8 m
sa
= 2
m s
-1 -2
s=
s=
20
mA
B
A
B u =
2 m
sa
= 4
m s
-1 -2
o
8 m
v=
A12
m s
-1
v=
B10
m s
-1
t =
0 s
t =
2 s
t =
4 s
mat
he
mat
ica
l c
al
cu
lat
ion
S
Dis
tanc
es a
nd sp
eeds
afte
r 2 s:
stt
vuat
vA A B
=+
=+
=+
= =+
=+
=+
=−
8 82
216
420 8
22
84
12
2
2
1
()
()
()
()(
)
m ms
==+
=+
=+
=−
uat
()
()(
)2
42
28
101
ms
st
ttt
st
tt
t
A B
=+
=+
=+
+
=+
+
()
()
()
()
82
8 82
4
82
2
1 22
2
1 22
2
Co
nt.
....
©The Dublin School of Grinds Page 20 Kieran Mills & Tony Kelly
ex
am
ple 3
Two
bodi
es m
ove
in th
e sa
me
dire
ctio
n al
ong
para
llel p
aths
. A st
arts
from
a p
oint
O w
ith v
eloc
ity 8
m s–1
and
ac
cele
ratio
n 2
m s–2
and
B st
arts
8 m
ahe
ad o
f A a
nd m
oves
off
with
vel
ocity
2 m
s–1 a
t acc
eler
atio
n 4
m s–2
. Fin
d w
hen
they
will
be
toge
ther
and
thei
r dis
tanc
es fr
om O
at t
hese
tim
es. W
hat a
re th
eir r
espe
ctiv
e sp
eeds
whe
n th
ey
are
toge
ther
?So
lu
tio
n
sut
at=
+1 2
2vuat
=+
st
ttt
st
tt
t
A B
=+
=+
=+
+
=+
+
()
()
()
()
82
8 82
4
82
2
1 22
2
1 22
2
mat
he
mat
ica
l c
al
cu
lat
ion
S
Dis
tanc
es a
nd sp
eeds
afte
r 4 s:
stt
vuat
vA A
=+
=+
=+
= =+
=+
=+
=−
8 84
432
1648 8
24
88
16
2
2
1
()
()
()
()(
)
m ms
BBuat
=+
=+
=+
=−
()
()(
)2
44
216
181
ms
u =
8 m
sa
= 2
m s
-1 -2
s=
s=
20
mA
B
A
B u =
2 m
sa
= 4
m s
-1 -2
o
8 m
v=
12
m s
A
-1v
= 1
6 m
sA
-1
v=
10
m s
B
-1v
= 1
8 m
sB
-1
s=
s=
48
mA
B
t =
0 s
t =
2 s
t =
4 s
©The Dublin School of Grinds Page 21 Kieran Mills & Tony Kelly
ex
am
ple 4
If A
star
ts 2
seco
nds b
efor
e B
find
whe
n an
d w
here
they
are
toge
ther
. Fin
d th
e m
axim
um d
ista
nce
that
A m
oves
ahe
ad o
f B in
the
subs
eque
nt m
otio
n.So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
sut
at=
+1 2
2
If A
star
ts 2
s be
fore
B, t
ake A
’s ti
me
as (t
+ 2
) s
and
B’s
tim
e as
t s.
u =
10
m s
a =
2 m
s
-1
-2
s=
sA
B
A
Bu =
1 m
sa
= 4
m s-1
-2
64
mt
t +
2
t =
0 s
t =
2 s
−+
−=
−+
=−
−=
=tt
tt
tt
t
2
2
1340
013
400
58
05
8(
)()
,ss
A a
nd B
are
toge
ther
(lev
el) 5
s an
d 8
s afte
r B st
arts
.
sepA
rAt
ion
eq
uAt
ion
: s
sA
B−
ss
tt
tt
tt
AB
−=
++
−−−
=−
+−
22
2
1424
264
1340
tog
eth
er (l
evel
): s
sA
B−
=0
Co
nt.
....
st
tt
tt
tt
t
A=
++
+
=+
++
=+
++
+
=+
()(
)(
)()
()
102
22
1020
210
204
41
1 22
2
2
244
24
641
4
264
1 22
2
t
st
tt
tB
+
=+
+
=+
+
()
()
st
tt
tt
tt
t
A=
++
+
=+
++
=+
++
+
=+
()(
)(
)()
()
102
22
1020
210
204
41
1 22
2
2
244
24
641
4
264
1 22
2
t
st
tt
tB
+
=+
+
=+
+
()
()
©The Dublin School of Grinds Page 22 Kieran Mills & Tony Kelly
ex
am
ple 4
If A
star
ts 2
seco
nds b
efor
e B
find
whe
n an
d w
here
they
are
toge
ther
. Fin
d th
e m
axim
um d
ista
nce
that
A m
oves
ahe
ad o
f B in
the
subs
eque
nt m
otio
n.So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
st
tA=
++
=+
+=
++
=
2
2
1424
814
824
6411
224
200
()
()
m
st
tA=
++
=+
+=
++
=
2
2
1424
514
524
2570
2411
9
()
()
m
Dis
tanc
e af
ter 5
s:
Dis
tanc
e af
ter 8
s:
sut
at=
+1 2
2
s=
s=
119
mA
B
A
B
s=
s=
20
0 m
AB
u =
10
m s
a =
2 m
s
-1
-2
u =
1 m
sa
= 4
m s-1
-2
64
mt
t +
2
t =
0 s
t =
5 s
t =
8 s
s=
t+
14t +
24
A
2
s=
2t
+ t +
64
B
2
(t +
2) =
2 s
(t +
2)=
7 s
(t +
2)=
10
s
Co
nt.
....
sepA
rAt
ion
eq
uAt
ion
: s
st
tA
B−
=−
+−
213
40
©The Dublin School of Grinds Page 23 Kieran Mills & Tony Kelly
ex
am
ple 4
If A
star
ts 2
seco
nds b
efor
e B
find
whe
n an
d w
here
they
are
toge
ther
. Fin
d th
e m
axim
um d
ista
nce
that
A m
oves
ahe
ad o
f B in
the
subs
eque
nt m
otio
n.So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
The
max
imum
sepa
ratio
n oc
curs
whe
n th
e ve
loci
ties o
f A
and
B a
re e
qual
.
Max
imum
Sep
arat
ion:
vv
AB
=
sut
at=
+1 2
2
s=
s=
119
mA
B
A
B
s=
s=
20
0 m
AB
u =
10
m s
a =
2 m
s
-1
-2
u =
1 m
sa
= 4
m s-1
-2
64
mt
t +
2
t =
0 s
t =
5 s
t =
8 s
s=
t+
14t +
24
A
2
s=
2t
+ t +
64
B
2
(t +
2) =
2 s
(t +
2)=
7 s
(t +
2)=
10
s
sepA
rAt
ion
eq
uAt
ion
: s
st
tA
B−
=−
+−
213
40
vuat
tt
t
vuatt
t
vv
A B AB
=+
=+
+=
++
=+
=+
=+
=+
=
102
210
24
214
14
41
2
()(
)
()
ttt
tt
ss
tt
AB
+=
+=
∴= −
=−
+−
=−
+−
=
144
113
2 65
1340
65
136
540
2
2
.
(.
)(
.)
s 2225.
m
©The Dublin School of Grinds Page 24 Kieran Mills & Tony Kelly
ExErcisE 4. catch up1. Two bodies start together at the same time at the same place and move along the same
straight line. If one moves with a constant speed of 8 m s–1 while the other starts from restand moves at a constant acceleration of 2 m s–2. How long will it take before they aretogether?
2. A car A passes a point P on a straight road at a constant speed of 10 m s–1. At the same timeanother car B starts from rest at P with uniform acceleration 2.5 m s–2.(i) When and how far from P will B overtake A.(ii) If B ceases to accelerate on overtaking, what time elapses between the two cars passing a
point Q which is 3 km from P.
3. A boy runs at 4 m s–1 away from a cyclist who starts at rest and accelerates at2 m s–2. If the boy has an initial lead of 5 m, how long does the cyclist take to catch him?
4. Two bodies A and B travel in the same direction along the same line. Body A starts withvelocity 3 m s–1 and acceleration 2 m s–2. The other body starts from the same place withvelocity 1 m s–1 and acceleration 3 m s–2. Find when and where they are together again.
5. Two bodies move along parallel tracks in the same direction. Body A starts with velocity2 m s–1 and acceleration 6 m s–2. Body B starts from the same place and the same time withvelocity 5 m s–1 and acceleration 2 m s–2. Find when and where they are together again. Findtheir velocities when they are together for the second time.
6. Two bodies move in the same direction along parallel paths. A starts from point O withvelocity 2 m s–1 and acceleration 4 m s–2. B starts 6 m ahead of A with velocity 3 m s–1 andacceleration 2 m s–2. Find when and where they are together and their velocities at thisinstant.
7. Two bodies move in the same direction along parallel paths. A starts from a point O withvelocity 8 m s–1 and acceleration 2 m s–2 and B starts 8 m ahead of A and moves off withvelocity 2 m s–1 at acceleration 4 m s–2. Find when they will be together and their distancesfrom O at these times.
8. Two bodies A and B travel in the same direction along the same line from the same point P atthe same time. A starts with velocity 5 m s–1 and acceleration 3 m s–2. B starts with velocity2 m s–1 and acceleration 4 m s–2. They are together again at point Q. Find the time at whichthey are together and the distance | PQ |. Find their maximum distance apart between P and Q.
9. Two bodies move in the same direction along parallel paths. They start at the same point P atthe same time. A starts from P with velocity 3 m s–1 and acceleration 2 m s–2. B starts withvelocity 1 m s–1 and acceleration 3 m s–2. They are together again at point Q. Find the time atwhich they are together and the distance | PQ |.Find their maximum distance apart between Pand Q.
10. Two bodies A and B move along parallel straight lines in the same direction from the samepoint P. A starts with velocity 4 m s–1 and acceleration 2 m s–2. B starts 1 second after A withvelocity 2 m s–1 and acceleration 4 m s–2. Find when and where they will be together.
©The Dublin School of Grinds Page 25 Kieran Mills & Tony Kelly
11. Two bodies A and B move along parallel straight lines in the same direction from the samepoint P. A starts from point P with velocity 5 m s–1 and acceleration 4 m s–2. B starts 1second before A with velocity 6 m s–1 and acceleration 3 m s–2 from a point a distance of2.5 m to the right of P. Find when and where they are together.
12. Find when and where they are together.Find their maximum separation between thetwo times when they are together.
13. IfAstarts2secondsbeforeBfindwhenandwhere they are together. Find their maximumseparation between the two positions.
14. A car A starts from a point P with initial velocity 8 m s–1 and then travels with uniformacceleration 4 m s–2. Two seconds later a second car B starts from P with an initial velocityof 30 m s–2 and then moves with a uniform acceleration of 3 m s–2. Show that after passing A,B will never be ahead by more than 74 m.
15. Bodies A and B start together and move along the same straight line. A starts with a speed of10 m s–1 and moves with a constant deceleration, while B starts at 5 m s–1 and accelerates at4 m s–2. Find the deceleration of A if they meet when the velocity of B is twice that of A.
16. The driver of a car travelling at 20 m s–1 sees a second car 120 m in front travelling in thesame direction at a uniform speed of 8 m s–1.(a) What is the least uniform retardation that must be applied to the faster car to avoid
collision?(b) If the actual retardation is 1 m s–2find
(i) the time interval in seconds for the faster car to reach a point 66 m behind the slowercar,
(ii) the shortest distance between the cars.
A
B
64 m
A
B
8 m
u = 10 m s–1 a = 2 m s–2
u = 1 m s–1 a = 4 m s–2
u = 10 m s–1 a = 2 m s–2
u = 1 m s–1 a = 4 m s–2
©The Dublin School of Grinds Page 26 Kieran Mills & Tony Kelly
answErs
ExErcisE 4
1. 8 s
2. (i) 8 s, 80 m (ii) 146 s
3. 5 s
4. 4 s, 28 m from starting point
5. 1.5 s, 9.75 m, 11 m s–1, 8 m s–1
6. 3 s, 24 m, 14 m s–1, 9 m s–1
7. 2 s, 4 s, 20 m, 48 m
8. 6 s, 84 m, 4.5 m
9. 4 s, 28 m, 2 m
10. 5 s after B starts, 60 m
11. 10 s after A starts, 250 m from P
12. 1 s, 8 s, 12.25 m
13. 5 s and 8 s after B starts, 119 m from A and 200 m from A, 2.25 m
15. 4 m s–2
16. (a) 0.6 m s–2
(b) (i) 6 s, 18 s; (ii) 48 m
©The Dublin School of Grinds Page 27 Kieran Mills & Tony Kelly
Sec
tio
n 4
: Ve
lo
cit
y-t
ime G
ra
phS
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly
dece
lera
tes u
nifo
rmly
to re
st fo
r 4 s.
Dra
w a
vel
ocity
-tim
e cu
rve
to re
p-re
sent
the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.
ex
am
ple 2
A c
ar tr
avel
s fro
m A
to B
. It s
tarts
from
rest
at A
and
acc
eler
ates
at
2 m
s–2 u
ntil
it re
ache
s a sp
eed
of 3
0 m
s–1. I
t the
n tra
vels
at t
his s
peed
fo
r 600
m a
nd th
en d
ecel
erat
es a
t 2.5
m s–2
to c
ome
to re
st a
t B. F
ind
(i)th
e to
tal t
ime
for t
he jo
urne
y,(ii
)th
e di
stan
ce fr
om A
to B
,(ii
i)th
e av
erag
e sp
eed
for t
he jo
urne
y.
ex
am
ple 3
A p
artic
le P
with
spee
d 14
0 m
s–1 b
egin
s to
dece
lera
te u
nifo
rmly
at a
ce
rtain
inst
ant w
hile
ano
ther
par
ticle
Q st
arts
from
rest
6 s
late
r and
ac-
cele
rate
s uni
form
ly. W
hen
the
seco
nd p
artic
le Q
has
trav
elle
d 12
5 m
, bo
th p
artic
les h
ave
a sp
eed
of 2
5 m
s–1.
(i)Sh
ow th
e m
otio
n of
bot
h on
the
sam
e sp
eed-
time
curv
e.(ii
)H
ow m
any
seco
nds a
fter t
he c
omm
ence
men
t of d
ecel
erat
ion
does
the
first
par
ticle
P c
ome
to re
st?
©The Dublin School of Grinds Page 28 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
8 6 416 14 12 10
20
24
28
32
48
1216
v (m
s)
-1
t (s
)
2 0
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n =
Slop
e of
cur
veD
ista
nce
= A
rea
unde
r cur
ve
co
nt...
.
©The Dublin School of Grinds Page 29 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
8
Run =
8 s
Ba
se
b =
8
14 12 10
20
24
28
32
48
1216
v (m
s)
-1
t (s
)
16
a = 2 ms
-2
6 4 26
4 m
0
Ris
e =
16
He
ight h
= 1
6m
s-1
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve c
on
t...
.
a sbh
==
=
==
=
−−
Ris
eR
unm
s sm
s
m
168
2
816
64
12
1 21 2(
)()
Are
a of
a tr
iang
le =
1 2bh
(i)
©The Dublin School of Grinds Page 30 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
8 6 4
Leng
th l
= 2
0
14 12 10
20
24
28
32
48
1216
v (m
s)
-1
t (s
)
216
a = 2 ms
-2
64
m
Bre
ad
th b
= 1
6
32
0 m
0
mat
he
mat
ica
l c
al
cu
lat
ion
S
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve c
on
t...
.
a sbh
==
=
==
=
−−
Ris
eR
unm
s sm
s
m
168
2
816
64
12
1 21 2(
)()
Are
a of
a re
ctan
gle
= lb
×
(i) (ii)slb
=×
==
()(
)20
1632
0m
©The Dublin School of Grinds Page 31 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
8 6 4
Run =
4 s
Ba
se
b =
4
14 12 10
20
24
28
32
48
1216
v (m
s)
-1
216
He
ight h
= 1
6
a = -4 m s-2
64
m3
20
m
a = 2 ms
-2
t (s
)3
2 m
Ris
e =
16
m s
-1
0
mat
he
mat
ica
l c
al
cu
lat
ion
S
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve
co
nt...
.
a sbh
==
=
==
=
−−
Ris
eR
unm
s sm
s
m
168
2
816
64
12
1 21 2(
)()
(i) (ii)slb
=×
==
()(
)20
1632
0m
Are
a of
a tr
iang
le =
1 2bh
(iii)a s
bh
==−
=−
==
=−−
Ris
eR
unm
s sm
s
m
164
4
416
3211
1 21 2(
)()
©The Dublin School of Grinds Page 32 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
co
nt...
.
(iv)
8 6 414 12 10
20
24
28
32
48
1216
v (m
s)
-1
216
t (s
)
a = -4 m s-2
64
m +
32
0 m
+ 3
2 m
= 4
16 m
a = 2 ms
-2
0
Ave
rage
Vel
ocity
Tota
l Dis
tanc
eTo
tal T
ime
=
Aver
age
velo
city
==
−41
6 3213
1m s
ms
©The Dublin School of Grinds Page 33 Kieran Mills & Tony Kelly
ex
am
ple 1
A c
ar st
artin
g fr
om re
st a
ccel
erat
es u
nifo
rmly
ove
r 8 s
to a
vel
ocity
of
16 m
s–1. I
t the
n m
aint
ains
a c
onst
ant v
eloc
ity fo
r the
nex
t 20
s. It
final
ly d
ecel
erat
es u
nifo
rmly
to re
st fo
r 4 s.
Dra
w
a ve
loci
ty-ti
me
curv
e to
repr
esen
t the
mot
ion
of th
e ca
r. U
se th
e gr
aph
to fi
nd(i)
the
acce
lera
tion
and
dist
ance
trav
elle
d by
the
car i
n th
e fir
st 8
s,(ii
)th
e di
stan
ce tr
avel
led
by th
e ca
r ove
r the
nex
t 20
s,(ii
i)th
e de
cele
ratio
n an
d di
stan
ce tr
avel
led
by th
e ca
r ove
r the
last
4 s,
(iv)
the
aver
age
velo
city
of t
he c
ar fo
r its
ent
ire jo
urne
y.So
lu
tio
n
8 6 414 12 10
20
24
28
32
48
1216
v (m
s)
-1
216
t (s
)
32
a = -4 m s-2
a = 2 ms
-2
20
16
0
416
m
mat
he
mat
ica
l c
al
cu
lat
ion
S
Are
a=
+1 2(
)xyh
Fin
din
g t
he
ar
ea u
nd
er t
he
cu
rve
in o
ne
go
:A
trap
eziu
m is
a fo
ur si
ded
shap
e w
here
two
of th
e si
des a
re p
aral
lel.
The
area
of a
trap
eziu
m is
hal
f the
su
m o
f the
par
alle
l sid
es b
y th
e pe
rpen
dicu
lar d
ista
nce
betw
een
them
.
s=
+
= = =
1 2 1 2
2032
1652
1652
841
6()(
)(
)()
()(
) m
©The Dublin School of Grinds Page 34 Kieran Mills & Tony Kelly
ex
am
ple 2
A c
ar tr
avel
s fro
m A
to B
. It s
tarts
from
rest
at A
and
acc
eler
ates
at
2 m
s–2 u
ntil
it re
ache
s a sp
eed
of 3
0 m
s–1. I
t the
n tra
vels
at t
his
spee
d fo
r 600
m a
nd th
en d
ecel
erat
es a
t 2.5
m s–2
to c
ome
to re
st
at B
. Fin
d(i)
the
tota
l tim
e fo
r the
jour
ney,
(ii)
the
dist
ance
from
A to
B,
(iii)
the
aver
age
spee
d fo
r the
jour
ney.
Sol
ut
ion
Tota
l tim
e T
= 15
s +
20 s
+ 12
s =
47 s
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve
mat
he
mat
ica
l c
al
cu
lat
ion
S
(i)I:aV t
t
t
=⇒
= ==
11
1230 30 2
15s
vs t
=s V t
= = =
−
600
301
2
m ms
?
II:V
s tt
t
=⇒
= ==
22
23060
0
600
3020
s
III:
.
.
aV t
t
t
=⇒
=
==
33
325
30
30 25
12s c
on
t....
.
©The Dublin School of Grinds Page 35 Kieran Mills & Tony Kelly
ex
am
ple 2
A c
ar tr
avel
s fro
m A
to B
. It s
tarts
from
rest
at A
and
acc
eler
ates
at
2 m
s–2 u
ntil
it re
ache
s a sp
eed
of 3
0 m
s–1. I
t the
n tra
vels
at t
his
spee
d fo
r 600
m a
nd th
en d
ecel
erat
es a
t 2.5
m s–2
to c
ome
to re
st a
t B
.Fin
d(i)
the
tota
l tim
e fo
r the
jour
ney,
(ii)
the
dist
ance
from
A to
B,
(iii)
the
aver
age
spee
d fo
r the
jour
ney.
Sol
ut
ion
mat
he
mat
ica
l c
al
cu
lat
ion
S
(ii)A
rea=
+1 2(
)xyh
S=
+=
1 220
4730
1005
()
m
Ave
rage
Vel
ocity
Tota
l Dis
tanc
eTo
tal T
ime
=
(iii)
==
−10
05 4721
41
m sm
s.
Aver
age
velo
city
Tota
l tim
e T
= 15
s +
20 s
+ 12
s =
47 s
©The Dublin School of Grinds Page 36 Kieran Mills & Tony Kelly
ex
am
ple 3
A p
artic
le P
with
spee
d 14
0 m
s–1 b
egin
s to
dece
lera
te u
nifo
rmly
at a
cer
tain
inst
ant w
hile
ano
ther
par
ticle
Q st
arts
fr
om re
st 6
s la
ter a
nd a
ccel
erat
es u
nifo
rmly
. Whe
n th
e se
cond
par
ticle
Q h
as tr
avel
led
125
m, b
oth
parti
cles
hav
e a
spee
d of
25
m s–1
.(i)
Show
the
mot
ion
of b
oth
on th
e sa
me
spee
d-tim
e cu
rve.
(ii)
How
man
y se
cond
s afte
r the
com
men
cem
ent o
f dec
eler
atio
n do
es t
he fi
rst p
artic
le P
com
e to
rest
?So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve
(ii) P
artic
le Q
:
125
255
101 21 2=
=
∴=
tt
t
()
s
sbh
=1 2
s h bt
= = =
−
125
251
m ms
©The Dublin School of Grinds Page 37 Kieran Mills & Tony Kelly
ex
am
ple 3
A p
artic
le P
with
spee
d 14
0 m
s–1 b
egin
s to
dece
lera
te u
nifo
rmly
at a
cer
tain
inst
ant w
hile
ano
ther
par
ticle
Q st
arts
fr
om re
st 6
s la
ter a
nd a
ccel
erat
es u
nifo
rmly
. Whe
n th
e se
cond
par
ticle
Q h
as tr
avel
led
125
m, b
oth
parti
cles
hav
e a
spee
d of
25
m s–1
.(i)
Show
the
mot
ion
of b
oth
on th
e sa
me
spee
d-tim
e cu
rve.
(ii)
How
man
y se
cond
s afte
r the
com
men
cem
ent o
f dec
eler
atio
n do
es t
he fi
rst p
artic
le P
com
e to
rest
?So
lu
tio
n
mat
he
mat
ica
l c
al
cu
lat
ion
S
(ii)P
artic
le P
:
Ve
lo
cit
y t
ime c
ur
Ve
S
Acc
eler
atio
n a
= Sl
ope
of c
urve
Dis
tanc
e s =
Are
a un
der c
urve
a=
=−
=−
−R
ise
Run
ms
115
167
182
.
u a v t
= =−
= =
−
−
−
140 718
0
1
2
1ms m
sm
s.
?
vuat
tt
t
=+
=+−
=
==
014
07
187
1814
014
07
1819
5
(.
).
..
s
©The Dublin School of Grinds Page 38 Kieran Mills & Tony Kelly
ExErcisE 5. VElocity timE curVEs
1. A car is travelling at 72 km h–1 when the brakes are applied producing a retardation of4 m s–2. How long does it take to stop?
2. An electric train starts from a station and reaches a speed of 14 m s–1 in 25 s with uniformacceleration.Sketchthevelocity-timegraph,andfindhowfarithasgonebythetimeitreaches this speed.
3. An aircraft can take off when it reaches a speed of 180 km h–1. If it attains this speed in 30 swith uniform acceleration what distance does it require for taking off?
4. An express train is travelling at 144 km h–1 when its brakes are applied. If these produce aretardation of 2 m s–2 how long will it take to stop and what distance will it cover in doingso?
5. A train starts from rest and attains a speed of 50 km h–1 in 4 minutes with uniformacceleration. It runs at that speed for 5 minutes and then slows down uniformly to rest in2minutes.Drawthevelocity-timegraphandfindthetotaldistancetravelled.
6. Find from the velocity-time graph shown(i) theaccelerationduringthefirst4s,(ii) the retardation during the last 2 s,(iii) the total distance travelled.
7. A cyclist rides along a straight road from A to B. He starts from rest at A and acceleratesuniformly to reach a speed of 10 m s–1 in 8 s. He maintains this speed for 30 s and thenuniformly decelerates to rest at B. If the total time is 48 s, draw a velocity-time curve andfromitfind(i) the acceleration,(ii) the deceleration,(iii) the total distance travelled.
8. A car travels from A to B. It starts from rest at A and accelerates at 1.5 m s–2 until it reaches aspeed of 30 m s–1. It then travels at this speed for 2 km and then decelerates at 2 m s–2 tocome to rest at B. Find(i) the total time for the journey,(ii) the distance from A to B,(iii) the average speed for the journey.
9. A and B are two points on a straight road. A car travelling along the road passes A whent = 0 and maintains a constant speed until t=20s,andinthistimecoversfour-fifthsofthedistance from A to B. The car then decelerates uniformly to rest at B. Draw a velocity-timecurveandfindthetimefromAtoB.
t (s)
v (m/s)
0 84 10
8
©The Dublin School of Grinds Page 39 Kieran Mills & Tony Kelly
10. A tram travels along a straight track and starts from rest. It accelerates uniformly for 20 sand during this time it travels 160 m. It maintains a constant speed for a further 50 s anddecelerates to rest in 8 s.Calculate(i) the acceleration,(ii) the deceleration,(iii) the total time,(iv) the total distance.
11. A train starts from rest and travels 8 km in 12 minutes ending at rest. The acceleration is halfthe retardation, both are uniform, and there is a period when the train runs at its maximumspeed of 50 km h–1. Find the time taken to reach full speed.
12. A 100 m sprinter starts with a speed of 6 m s-1 and accelerates uniformly to 10 m s–1 andfinishestheraceatthisspeed.Ifhistotaltimeis10.4s,findhisuniformaccelerationandafter what distance he is going at full speed.
13. Acartakes2minutestotravelbetweentwosetsoftrafficlights2145mapart.Ithasuniformacceleration for 30 s, then uniform velocity, and then uniform retardation for the last 15 s.Find the maximum velocity and the acceleration.
14. A train travels 15 km between two stations at an average speed of 50 km h–1. Its accelerationis half the retardation and both are uniform. If the maximum speed is 72 km h–1findtheacceleration in m s–2. Sketch the velocity-time curve.
15. A car accelerates at 2 m s–2 in bottom gear, 1.5 m s–2 in second gear and 1 m s–2 in top gear.Each gear change takes 1.5 s during which time the car travels at constant speed. If amotorist changes gear when his speeds are 3 m s–1 and 9 m s–1findhowlonghewilltaketoreach 15 m s–1 from rest.
16. A train moving in a straight line starts from A with uniform acceleration of 0.1 m s–2. After ithas attained full speed it moves uniformly for 10 minutes. It is brought to rest at B by thebrakes, which apply a constant retardation of 0.8 m s–2 for 20 s. Draw a rough velocity-timegraphandfromitfindthetimeofthejourneyandthedistancefromA to B.
17. A train has a maximum speed of 72 km h–1 which it can achieve at an acceleration of0.25 m s–2. With its brakes fully applied the train has a deceleration of 0.5 m s–2. What is theshortest time that the train can travel between stations 8 km apart if it stops at both stations?
18. A particle with speed 150 m s–1 begins to decelerate uniformly at a certain instant whileanother particle starts from rest 8 s later and accelerates uniformly. When the second particlehas travelled 135 m, both particles have a speed of 30 m s–1.(i) Show the motion of both on the same speed-time curve.(ii) Howmanysecondsafterthecommencementofdecelerationdoesthefirstparticlecome
to rest?
©The Dublin School of Grinds Page 40 Kieran Mills & Tony Kelly
19. A body starts from rest at P travelling in a straight line and then comes to rest at Q which is696 m from P.Thetimetakenis66s.Forthefirst10sithasuniformaccelerationa. It thentravelsatconstantspeedandisfinallybroughttorestbyauniformdecelerationb acting for6 s. Find a and b.If the journey from rest at P to rest at Q had been travelled with no interval of constant speedbut at acceleration of a for a time t1 immediately followed by deceleration b for a time t2,show that the time for the journey is 8 29 s.
20. An athlete runs 100 m in 12 s. Starting from rest he accelerates uniformly to a speed of10 m s–1 and then continues at that speed. Calculate the acceleration.
21. A cyclist has a maximum acceleration of 2 m s–2, a maximum speed of 15 m s–1 and amaximum deceleration of 4 m s–2. If he travels from rest to rest in the shortest possible timeshow that he covers a distance of 84 3
8 m. Find the time to travel(i) 105 m, (ii) 54 m.
AnswErs
ExErcisE 5
1. 5 s
2. 175 m
3. 0.75 km
4. 20 s, 400 m
5. 6 23 km
6. (i) 2 m s–2 (ii) 4 m s–2 (iii) 56 m
7. (i) 1.25 m s–2 (ii) 1 m s–2 (iii) 390 m
8. (i) 3053 s (ii) 2525 m (iii) 24.83 m s–1
9. 30 s
10. (i) 0.8 m s–2 (ii) 2 m s–2
(iii) 78 s (iv) 1024 m
11. 3.2 minutes
12. 2 m s–2, 16 m
13. 22 m s–1, 0.73 m s–2
14. 122
2ms−
15. 14.5 s
16. 13 minutes, 11.04 km
17. 460 s
18. (ii) 21.25 s
19. 1.2 m s–2, 2 m s–2
20. 2.5 m s–2
21. (i) 12 58 s (ii) 9 s
©The Dublin School of Grinds Page 41 Kieran Mills & Tony Kelly
Sec
tio
n 5
: Fr
ee F
al
l
ex
am
ple 1
A b
ody
is th
row
n ve
rtica
lly u
p fr
om th
e gr
ound
at 1
4 m
s-1 .
Find
the
max
imum
hei
ght i
t rea
ches
and
the
time
to re
ach
this
hei
ght.
From
its h
ighe
st p
oint
find
the
time
for t
he b
ody
to h
it th
e gr
ound
and
its
spee
d w
hen
it hi
ts th
e gr
ound
.
ex
am
ple 2
A b
all i
s thr
own
verti
cally
up
at 1
0 m
s-1 f
rom
a p
oint
3 m
abo
ve th
e gr
ound
. Fin
d th
e sp
eed
it ha
s whe
n it
hits
the
grou
nd a
nd th
e tim
e it
take
s the
bal
l to
hit t
he g
roun
d.
ex
am
ple 3
A h
ot-a
ir ba
lloon
trav
els f
rom
the
grou
nd v
ertic
ally
up
at a
con
stan
t sp
eed
of 1
2 m
s-1 .
Find
its h
eigh
t abo
ve th
e gr
ound
afte
r 5 s.
A
fter 5
s a
ball
is d
ropp
ed fr
om th
e ba
loon
. How
long
doe
s it t
ake
to
reac
h th
e gr
ound
?
©The Dublin School of Grinds Page 42 Kieran Mills & Tony Kelly
ex
am
ple 1
A b
ody
is th
row
n ve
rtica
lly u
p fr
om th
e gr
ound
at 1
4 m
s-1 .
Find
the
max
imum
hei
ght i
t rea
ches
and
the
time
to re
ach
this
hei
ght.
From
its h
ighe
st p
oint
find
its s
peed
whe
n it
hits
the
grou
nd a
nd th
e tim
e fo
r the
bod
y to
hit
the
grou
nd.
Sol
ut
ion
v=
0 m
s�1
s=
?
u=
14 m
s�1
GO
IN
GU
P
The
acce
lera
tion
due
to g
ravi
ty is
den
oted
by
g.Th
is v
alue
is 9
.8 m
s-2 .
In fr
ee fa
ll al
way
let a
= -
g = -9
.8 m
s-2 .
Go
inG
up:
u =
14 m
s-1
v =
0 m
s-1
a = -9
.8 m
s-1
s = ?
t = ?
vu
ass
s
s22
22
20
142
98
196
196
196
196
10
=+
=+
−=
∴=
=(.
).
.m
vuat
tt
t
=+
=−
=
==
=
014
98
98
1414 9
810 7
143
..
..
s
Go
inG
do
wn
:u
= 0
m s-
1
a = -9
.8 m
s-1
s = -
10 m
t = ?
v =
?
vu
asv v
v22
22
2
1
20
29
810
196 19
614
=+
=+
−−
=
∴=
=−
−
(.
)()
ms
vuat
t
t
=+
−=
−
==
=
140
98
14 98
10 71
43
.
..
s
Up
is p
ositi
veD
own
is n
egat
ive
Up
is p
ositi
veD
own
is n
egat
ive
u=
0 m
s�1
s=
10 m
�
v=
?
GO
IN
GD
OW
N
co
nc
lu
Sio
nS:
Tim
e up
= T
ime
Dow
n
Velo
city
goi
ng u
p =
Velo
city
at s
ame
poin
t on
way
dow
n bu
t in
the
oppo
site
dire
ctio
n
©The Dublin School of Grinds Page 43 Kieran Mills & Tony Kelly
ex
am
ple 2
A b
all i
s thr
own
verti
cally
up
at 1
0 m
s-1 f
rom
a p
oint
3 m
abo
ve th
e gr
ound
. Fin
d th
e sp
eed
it ha
s whe
n it
hits
the
grou
nd a
nd th
e tim
e it
take
s the
bal
l to
hit t
he g
roun
d.So
lu
tio
n
u=
10 m
s�1
s=
3 m
�
v=
?
u =
10 m
s-1
s = -
3 m
a = -9
.8 m
s-1
v =
?t =
?
vu
asv v2
2
22 2
1
210
29
83
102
98
312
6
=+
=+
−−
=+
−−
=−
−
(.
)()
(.
)()
.m
s
vuat
tt
t
=+
−=
−−
=−
==
126
109
822
69
822
69
82
31
..
..
. ..
sTh
e ac
cele
ratio
n du
e to
gra
vity
is d
enot
ed b
y g.
This
val
ue is
9.8
m s-
2 .In
free
fall
alw
ay le
t a =
-g
= -9
.8 m
s-2 .
Up
is p
ositi
veD
own
is n
egat
ive
The
velo
city
is n
egat
ive
as it
is m
ovin
g do
wn.
©The Dublin School of Grinds Page 44 Kieran Mills & Tony Kelly
ex
am
ple 3
A h
ot-a
ir ba
lloon
trav
els f
rom
the
grou
nd v
ertic
ally
up
at a
con
stan
t spe
ed o
f 12
m s-
1 . Fi
nd it
s hei
ght a
bove
the
grou
nd a
fter 5
s.
Afte
r 5 s
a ba
ll is
dro
pped
from
the
ballo
on. H
ow lo
ng d
oes i
t tak
e to
reac
h th
e gr
ound
? G
ive
your
ans
wer
to o
ne
plac
e of
dec
imal
.So
lu
tio
n
vs t
=v t vs t
svt
= = =⇒
=×
=×
=
−12 5
125
60
1m
ss
m
Bal
l is d
ropp
ed fr
om b
allo
on:
u =
12 m
s-1
s = -
60 m
a = -9
.8 m
s-2
t = ?
u=
12 m
s�1
t=
5 s
s=
?v
= 1
2 m
s�1
v=
12 m
s�1
Bal
l is
dro
pped
s=
60 m
�
t=
?
sut
at tt
tt
tt
t=+
−=
+−
−=
−
−−
=
∴=
1 22 1 2
2
2
2
6012
98
6012
49
49
1260
0
(.
)
..
11212
44
960
24
94
92
±−
−−
=(
)(
.)(
)(
.)
.s
©The Dublin School of Grinds Page 45 Kieran Mills & Tony Kelly
ExErcisE 6. Uniformly g accElEratEd motion
[In all problems g = 9.8 m s–2]
1. A vase falls from a shelf 140 cm above the floor. Find the speed with which it strikes thefloor.
2. A stone is dropped from a point 49 m above the ground. Find the time for it to reach theground.
3. A stone is thrown down at 5 m s–1. If its speed on hitting the ground is 19 m s–1 from whatheight was it thrown. How long does it take?
4. A stone is dropped from the top of a tower and falls to the ground. If it strikes the ground at14 m s–1, how high is the tower?
5. A ball is thrown vertically downwards from the top of a tower with an initial speed of2 m s–1. If it hits the ground 3 s later find(i) the height of the tower,(ii) the speed with which it hits the ground.
6. A stone is thrown upwards with a speed of 21 m s–1. Find its height(i) 1 s after projection,(ii) 2 s after projection,(iii) 3 s after projection.
7. A ball is thrown up at 14 m s–1 from a point 2 m above the ground. Find(i) the speed when it returns to the level of projection,(ii) the speed on the ground.
8. A ball is thrown vertically up at 28 m s–1. Find(i) the maximum height,(ii) the time to reach the maximum height,(iii) the velocity of return,(iv) the total time for the journey.
9. A balloon is rising at a steady speed of 3 m s–1. How high is it above the ground after 10 s?At this instant a man releases a stone. What is the initial velocity of the stone?How long does it take to reach the ground? How high is the balloon above the ground whenthe stone strikes the ground?
10. A stone is thrown up at 49 m s–1 from the ground. Find the times at which the particle is78.4 m above the ground. Find the time interval for which the particle is above 78.4 m.
11. A ball is thrown up at 14 m s–1. Find the times at which the particle is 9.1 m above theground.
©The Dublin School of Grinds Page 46 Kieran Mills & Tony Kelly
12. A ball is thrown up at 49 m s–1. How long does it take to reach its maximum height? If another ball was thrown up 1 s after the first one, how high is it above the ground when the firstball has reached its maximum height if it has the same initial velocity?
13. A jumper can jump 2 m on the Earth. What is his take-off speed? How high can he jump onthe moon? (Acceleration due to gravity of moon g = 1.6 m s–1 )
14. A particle is thrown vertically upwards under gravity with a speed of 16 m s–1. One secondlater another particle is fired upwards from the same point. Find the initial speed of thisparticle in order that the two particles will collide when the first particle has reached itshighest point.
15. An object falls vertically past a window 2 m high in 112 s. Find the height above the window
from which the object was dropped.
16. A stone is dropped from a balloon rising at 10 m s–1 and reaches the ground in8 s. How high was the balloon above the ground when the stone was dropped?
17. A body falls from the top of a tower and during the last second it falls 925 of the total
distance. Find the height of the tower.
18. A particle falls freely from rest from a point O passing three points A, B and C, the distances|AB| and |BC| being equal. If the particle takes 3 s to pass from A to B and 2 s from B to C,calculate |AB|.
19. A body falls freely from rest from a point O passing three points A, B and C, the distances|AB| and |BC| being equal. The time taken to go from A to B is 2 s and from B to C is 1 s.Find |AB|.
20. A particle falls freely under gravity from rest at a point P. After it has fallen for 1 s anotherparticle is projected vertically downwards from P with speed 14.7 m s–1. Find the time anddistance from P at which they collide.
©The Dublin School of Grinds Page 47 Kieran Mills & Tony Kelly
Answers
exercise 6
1. 5.24 m s–1
2. 3.16 s
3. 17.14 m, 1.43 s
4. 10 m
5. (i) 50.1 m (ii) 31.4 m s–1
6. (i) 16.1 m (ii) 22.4 m (iii) 18.9 m
7. (i) 14 m s–1 (ii) 15.34 m s–1
8. (i) 40 m (ii) 207 s (iii) 28 m s–1
(iv) 407 s
9. 30 m, 3 m s–1, 2.8 s, 38.4 m
10. 2 s, 8 s, 6 s
11. 1 s, 137 s
12. 5 s, 117.6 m
13. 6.26 m s–1, 12.25 m
14. 23.7 m s–1
15. 28.4 m
16. 233.6 m
17. 122.5 m
18. 147 m
19. 29.4 m
20. 2 s, 19.6 m
©The Dublin School of Grinds Page 48 Kieran Mills & Tony Kelly
20131. (a) A ball is thrown vertically upwards with a speed of 44·1 m s−1.
Calculate the time interval between the instants that the ball is 39·2 m above the pointof projection.
(b) A lift ascends from rest with constant acceleration f until it reaches a speed v. Itcontinues at this speed for t1 seconds and then decelerates uniformly to rest withdeceleration f.The total distance ascended is d, and the total time taken is t seconds.
(i) Draw a speed-time graph for the motion of the lift.
(ii) Show that v f t t= -12 1( ).
(iii) Show that t t df1
2 4= − .
LeAving cert Questions20141. (a) Two cars, P and Q, travel with the same constant velocity 15 m s–1 along a straight level
road. The front of car P is 24 m behind the rear of car Q. At a given instant both cars decelerate, P at 4 m s–2 and Q at 5 m s–2.
(i) Find, in terms of t, the distance between the cars t seconds later.
(ii) Find the distance between the cars when they are at rest.
©The Dublin School of Grinds Page 49 Kieran Mills & Tony Kelly
20121. (a) A particle falls from rest from a point P. When it has fallen a distance 19·6 m a second
particle is projected vertically downwards from P with initial velocity 39·2 m s–1.The particles collide at a distance d from P.Find the value of d.
(b) A car, starts from rest at A, and accelerates uniformly at 1 m s–2 along a straight level roadtowards B, where AB = 1914 m. When the car reaches its maximum speed of 32 m s–1, itcontinues at this speed for the rest of the journey.At the same time as the car starts from A a bus passes B travelling towards A with aconstant speed of 36 m s–1. Twelve seconds later the bus starts to decelerate uniformly at0·75 m s–2.
(i) The car and the bus meet after t seconds. Find the value of t.
(ii) Find the distance between the car and the bus after 48 seconds.
20111. (a) A particle is released from rest at A and falls vertically passing
two points B and C.
It reaches B after t seconds and takes 27t
seconds to fall from B to C, a distance of 2.45 m.
Find the value of t.
(b) A car accelerates uniformly from rest to a speed v in t1 seconds.It continues at this constant speed for t seconds and then deceleratesuniformly to rest in t2 seconds.
The average speed for the journey is 34v .
(i) Draw a speed-time graph for the motion of the car.
(ii) Find t1 + t2 in terms of t.
(iii) If a speed limit of 23v were to be applied, find in terms of t the least time the
journey would have taken, assuming the same acceleration and deceleration as inpart (ii).
A
B
C
©The Dublin School of Grinds Page 50 Kieran Mills & Tony Kelly
20101. (a) A car is travelling at a uniform speed of 14 ms–1 when the driver notices a traffic light
turning red 98 m ahead.
Find the minimum constant deceleration required to stop the car at the traffic light,(i) if the driver immediately applies the brake(ii) if the driver hesitates for 1 second before applying the brake.
(b) A particle passes P with speed 20 ms–1 and moves in a straight line to Q with uniformacceleration.In the first second of its motion after passing P it travels 25 m.In the last 3 seconds of its motion before reaching Q it travels 13
20 of |PQ|.
Find the distance from P to Q.
20091. (a) A particle is projected vertically upwards from
the point P. At the same instant a second particleis let fall vertically from Q.The particles meet at R after 2 seconds.
The particles have equal speeds when they meetat R.
Prove that |PR| = 3|RQ|.
(b) A train accelerates uniformly from rest to a speed v m/s with uniformacceleration f m/s2.
It then decelerates uniformly to rest with uniform retardation 2f m/s2.The total distance travelled is d metres.(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is d3
, find the value of f.
Q
R
P
©The Dublin School of Grinds Page 51 Kieran Mills & Tony Kelly
20081. (a) A ball is thrown vertically upwards with an initial velocity of 39.2 m/s. Find
(i) the time taken to reach the maximum height
(ii) the distance travelled in 5 seconds.
(b) Two particles P and Q, each having constant acceleration, are moving in the samedirection along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s,respectively. Two minutes later Q passes P, and Q is then moving at 65.5 m/s. Find(i) the acceleration of P and the acceleration of Q
(ii) the speed of P when Q overtakes it
(iii) the distance P is ahead of Q when they are moving with equal speeds.
20071. (a) A particle is projected vertically downwards from the top of a
tower with speed u m/s. It takes the particle 4 seconds to reach the bottom of the tower.During the third second of its motion the particle travels 29.9 metres.
Find(i) the value of u
(ii) the height of the tower.
(b) A train accelerates uniformly from rest to a speed v m/s.It continues at this speed for a period of time and then decelerates uniformly to rest.In travelling a total distance d metres the train accelerates through a distancepd metres and decelerates through a distance qd metres, where p < 1 and q < 1.(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is vp q b+ +
,
find the value of b.
29.9 m
t = 0
t = 2
t = 3
©The Dublin School of Grinds Page 52 Kieran Mills & Tony Kelly
20061. (a) A lift starts from rest. For the first part of its descent it travels with uniform
acceleration f. It then travels with uniform retardation 3f and comes to rest.The total distance travelled is d and the total time taken is t.(i) Draw a speed-time graph for the motion.
(ii) Find d in terms of f and t.
(b) Two trains P and Q, each of length 79.5 m, moving in opposite directions along parallellines, meet at O, when their speeds are 15 m/s and 10 m/s respectively.The acceleration of P is 0.3 m/s2 and the acceleration of Q is 0.2 m/s2.It takes the trains t seconds to pass each other.(i) Find the distance travelled by each train in t seconds.
(ii) Hence, or otherwise, calculate the value of t.
(iii) How long does it take for 25 of the length of train Q to pass the point O?
20051. (a) Car A and car B travel in the same direction along a horizontal straight road.
Each car is travelling at a uniform speed of 20 m/s.Car A is at a distance of d metres in front of car B.At a certain instant car A starts to brake with a constant retardation of 6 m/s2.0.5 s later car B starts to brake with a constant retardation of 3 m/s2 .Find(i) the distance travelled by car A before it comes to rest
(ii) the minimum value of d for car B not to collide with car A.
20041. (a) A ball is thrown vertically upwards with an initial velocity of 20 m/s.
One second later, another ball is thrown vertically upwards from the same point with an initial velocity of u m/s.The balls collide after a further 2 seconds.(i) Show that u = 17.75.
(ii) Find the distance travelled by each ball before the collision, giving your answerscorrect to the nearest metre.
©The Dublin School of Grinds Page 53 Kieran Mills & Tony Kelly
20031. (a) The points P, Q and R all lie in a straight line.
A train passes point P with speed u m/s. The train is travelling with uniform retardation f m/s2. The train takes 10 seconds to travel from P to Q and 15 seconds to travel from Q to R, where | PQ | = | QR | = 125 metres.
(i) Show that f = 13 .
(ii) The train comes to rest s metres after passing R.Find s, giving your answer correct to the nearest metre.
(b) A man runs at constant speed to catch a bus.At the instant the man is 40 metres from the bus, it begins to accelerate uniformly fromrest away from him.The man just catches the bus 20 seconds later.(i) Find the constant speed of the man.
(ii) If the constant speed of the man had instead been 3 m/s, show that the closest he getsto the bus is 17.5 metres.
20021. (a) A stone is thrown vertically upwards under gravity with a speed of u m/s from a point
30 metres above the horizontal ground. The stone hits the ground 5 seconds later. (i) Find the value of u.
(ii) Find the speed with which the stone hits the ground.
(b) A particle, with initial speed u, moves in a straight line with constant acceleration.During the time interval from 0 to t, the particle travels a distance p.During the time interval from t to 2t, the particle travels a distance q.During the time interval from 2t to 3t, the particle travels a distance r.(i) Show that 2q = p + r.
(ii) Show that the particle travels a further distance 2r – q in the time interval from 3tto 4t.
©The Dublin School of Grinds Page 54 Kieran Mills & Tony Kelly
20011. (a) Points P and Q lie in a straight line, where | PQ | = 1200 metres.
Starting from rest at P, a train accelerates at 1 m/s2 until it reaches the speed limit of 20 m/s. It continues at this speed of 20 m/s and then decelerates at 2 m/s2, coming to rest at Q. Find the time it takes the train to go from P to Q. Find the shortest time it takes the train to go from rest at P to rest at Q if there is no speed limit, assuming that the acceleration and deceleration remain unchanged at 1 m/s2
and 2 m/s2, respectively.
(b) A particle is projected vertically upwards with an initial velocity of u m/s andanother particle is projected vertically upwards from the same point and with the sameinitial velocity T seconds later.Show that the particles
(i) will meetT u
g2+
seconds from the instant of projection of the first particle
(ii) will meet at a height of4
8
2 2 2u g Tg−
metres.
20001. (a) A stone projected vertically upwards with an initial speed of u m/s rises 70 m in the first
t seconds and another 50 m in the next t seconds.Find the value of u.
(b) A car, starting from rest and travelling from P to Q on a straight level road, where| PQ | = 10 000 m, reaches its maximum speed 25 m/s by constant acceleration in the first500 m and continues at this maximum speed for the rest of the journey.A second car, starting from rest and travelling from Q to P, reaches the samemaximum speed by constant acceleration in the first 250 m and continues at thismaximum speed for the rest of the journey.(i) If the two cars start at the same time, after how many seconds do the two cars meet?
Find, also, the distance travelled by each car in that time.
(ii) If the start of one car is delayed so that they meet each other exactly halfwaybetween P and Q, find which car is delayed and by how many seconds.
19991 (b) A particle travels in a straight line with constant acceleration f for 2t seconds and covers
15 metres. The particle then travels a further 55 metres at constant speed in 5t seconds. Finally the particle is brought to rest by a constant retardation 3f.
(i) Draw a speed-time graph for the motion of the particle.
(ii) Find the initial velocity of the particle in terms of t.
(iii) Find the total distance travelled in metres, correct to two decimal places.
©The Dublin School of Grinds Page 55 Kieran Mills & Tony Kelly
19981 (a) A train accelerates uniformly from rest to a speed v m/s. It continues at this constant
speed for a period of time and then decelerates uniformly to rest. If the average speed for the whole journey is 5
6 v, find what fraction of the whole distance is described at constant speed.
(b) Car A, moving with uniform acceleration 320 b m/s2 passes a point P with speed 9u m/s.
Three seconds later car B, moving with uniform acceleration 29 b m/s2 passes the same
point with speed 5u m/s. B overtakes A when their speeds are 6.5 m/s and 5.4 m/s respectively. Find(i) the value u and the value b,
(ii) the distance travelled from P until overtaking occurs.
19971 (a) A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It
continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance travelled while accelerating is 6 m. The total distance travelled is 30 m and the total time taken is 6 s.
(i) Draw a speed-time graph and hence, or otherwise, find the value of v.
(ii) Calculate the distance travelled at v m/s.
(b) Two points P and Q are a distance d apart. A particle starts from P and move towards Qwith initial velocity 2u and uniform acceleration f. A second particle starts at the sametime from Q and moves towards P with initial velocity 3u and uniform deceleration f.Prove that
(i) the particles collide afterdu5
seconds,
(ii) if the particles collide before the second particle comes to instantaneous rest, thenfd u<15 2,
(iii) if fd u= 30 2 then the second particle has returned to Q before the collision.
©The Dublin School of Grinds Page 56 Kieran Mills & Tony Kelly
19961 (a) A particle starts from rest and moves in a straight line with uniform acceleration. It
passes three points A, B and C where |AB| = 105 m and |BC| = 63 m. If it takes 6 seconds to travel from A to B and 2 seconds to travel from B to C find(i) its acceleration
(ii) the distance of A from the starting position.
(b) A lift starts from rest with constant acceleration 4 m/s2. It then travels with uniformspeed and finally comes to rest with constant retardation 4 m/s2. The total distancetravelled is d and the total time taken is t.(i) Draw a speed-time graph.
(ii) Show that the time for which it travelled with uniform speed is t d2 − .
©The Dublin School of Grinds Page 57 Kieran Mills & Tony Kelly
Answers
LeAving cert. Questions
2014. 1 (a) (i) 24 12
2- t , (ii) 18.375 m
2013. 1 (a) 7 s
2012. 1 (a) 44.1 m(b) (i) 40 s, (ii) 352 m
2011. 1 (a) t = 78 s
(b) (ii) t = t1 + t2, (iii) 3112 t
2010. 1 (a) (i) –1 m s–2, (ii) − −76
2m s(b) 300 m
2009. 1 (b) (ii) f = 1 m s–2
2008. 1 (a) (i) 4 s, (ii) 83.3 m(b) (i) a aP Q= =5
242 20 5m/s m/s, . (ii) 48 m/s (iii) 525 m
2007. 1 (a) (i) 5.4 m/s, (ii) 100 m; (b) (ii) b = 1
2006. 1 (a) (ii) 38
2ft ;
(b) (i) s t t s t tP Q= + = +15 0 15 10 0 12 2. , . , (ii) 6 s, (iii) 3.1 s
2005. 1 (a) (i) 1003 m, (ii) 43.3 m
2004. 1 (a) (ii) 25 m, 16 m
2003. 1 (a) (ii) 51 m; (b) (i) 4 m/s
2002. 1 (a) (i) 18.5 m/s, (ii) 30.5 m/s;
2001. 1 (a) 75 s, 60 s;
2000. 1 (a) 56 m/s;(b) (i) 215 s, P: 4875 m, Q: 5125 m, (ii) Delay car starting at Q by 5 s
1999. 1 (b) (ii) 4t
, (iii) 75.76 m
1998. 1 (a) 45
(b) (i) u = 0.1 m/s, b = 1, (ii) 94.5 m
1997. 1 (a) (i) 6.5 m/s, (ii) 21 m;
1996. 1 (a) (i) 3.5 m/s2, (ii) 7 m
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