(e) Displacement-time graph and velocity

(i) The slope of the displacement-time graph represents the velocity of the body.

s / m

t / s

The graph is a hoirzontal line. Velocity is zero.

s / m

t / s

The slope > 0 (positive). The object is moving towards the end position

s / m

t / s

The slope <0 (negative). The object is moving towards the start position

IX. Speed-Time Graph and Velocity-Time Graph

5 cm 6 cm 4 cm 2 cm< >>><><<

t = 0s = 0

The average speed or velocity of the four portions are :v1 = 0.05 m / 0.1 s = 0.5 ms-1

v2 = 0.06 m / 0.1 s = 0.6 ms-1

v3 = 0.04 m / 0.1 s = 0.4 ms-1

v4 = 0.02 m / 0.1 s = 0.2 ms-1

Speed or velocity is at middle instant of the time interval.

t / s0.1 0.2 0.3 0.4

0.2

0.4

0.6

0

speed / ms -1

Constructing Speed time graph or velocity time graph

constructed by sticking the ticker-tape

0.1 0.2 0.3 0.4t / s

speed / ms-1

0

0.2

0.4

0.6

t / s

distance / m

0.1 0.2 0.3 0.4

0.05

0.10

0.15

0

(1)

(2)

(3) (4)

portion (1) , from 0 s to 0.1 s ,m1 = 0.04 m / 0.1 s = 0.4 ms-1

portion (2) , from 0.1 s to 0.2 s ,m2 = 0.06 m / 0.1 s = 0.6 ms-1

portion (3) , from 0.2 s to 0.3 s ,m3 = 0.02 m / 0.1 s = 0.2 ms-1

portion (4) , from 0.3 s to 0.4 s ,m4 = 0.02 m / 0.1 s = 0.2 ms-1

Construct from a distance-time graph (or a displacement-time graph)

Displacement and v-t graph

• The following slides show you the relationship between displacement (S) and the v-t graph (velocity-time graph)

t / s

speed / ms

0.1 0.2 0.3 0.4

0.20

0.40

0.60

0

(1)

(2)

(3) (4)

-1

Area = 0.6 x (0.2- 0.1)

Area = speed x time = distance

Graph area under the graph?

area under the graph from t = 0.1s to 0.2s is the distance travelled during t=0.1s to 0.2s

v / ms

t / s

-1

V1

A

t t1 2

moves with a fixed (constant) velocity.

= v1(t2 - t1)Displacement travelled by the object from time t1 to t2 = area A

The area under v-t graph

V/ms-1

t/s

v2

v1

t2t1

Displacement due to v1 from t1 to t2

Displacement due to the increase of velocity from v1 to v2 from t1 to t2

The slope of the velocity-time graph???

X. Acceleration ( )加速度

(a) It is a vector quantity(can be ‘+’ or ‘-‘)(b) the rate of change of velocity

(c) Unit : ms-2

(d) Symbol : a

,interval time

yin velocit changeon = accelerati

ta

v

= or

t / s

v / ms-1

The slope is positive

The acceleration is positive.

The velocity is increasing.

v / ms-1

t / s

The slope is negative

The acceleration is negative.

The velocity is decreasing.

Case 1 : Initial velocity = 0 (zero initial velocity)

Displacement vs Time

0

1

2

3

4

5

6

7

8

9

0 2 4 6 8 10time/s

Dis

plac

emen

t/m

Velocity vs Time

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 1 2 3 4 5 6 7 8 9time/s

Vel

oci

ty/m

s-1

(i) What kind of motion does the v-t graph represent?The velocity is increasing, uniform acceleration

(ii) What relationship best describes the motion ?v is directly proportional to t

(iii) Write down the equation that relates to the v-t graph.

(iv) What is the meaning of the slope m? change in velocity per second, a = (1/5) ms-2

(v) The area covered in the v-t graph within the first 8 seconds? Same as the distance? S = 6.4 m

c is the y-intercept, c = 0. Slope m= a. The equationis v = at y = mx

(vi) Write down anequation to relatethe displacement, S,in terms of velocity,v, and time, t?

Area of Triangle forthe first t seconds,S = v t / 2S = 1/2 a t2

Velocity vs Time

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 1 2 3 4 5 6 7 8 9time/s

Vel

oci

ty/m

s-1

‘at’ or

‘v’

t

S = area

Velocity vs Time

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9 10time/s

Ve

loc

ity

/ms

-1

Case 2 : Initial velocity> 0(i) What kind of motion does

the v-t graph represent?

Uniform acceleration(ii) What relationship best

describes the motion?

Linear relationship(iii) Write down an equation

in the form of y=mx + c?

c is the y-intercept,v = a t + u, or v = u + a ty mx c

Velocity vs Time

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9 10time/s

Ve

loc

ity

/ms

-1

(iv) What is the area covered by thev-t graph in the first 8 seconds?

6.4 + 1.2 x 8 = 16m

(V) Write down an equation torelate the displacement interms of velocity (u) andtime (t)

Area of Triangle for the firstt seconds, S= u t + 1/2 (a t2)

S = (v+u) t /2ORu ut

at

t

1/2 (a t2)

u

v

t

S=(u+v)t/2

XI.Equations of uniform accelerated motion1. v = u + a t…………. (1)

2. s = u t + ½ a t 2 …… (2)

3. s = (v+u) t /2………. (3)

From (1) t = (v-u)/a & (3)

s = (v+u) (v-u)/(2a)

4. 2 a s = v2 – u2……….. (4)

no sno vno a

no t

Where u = Initial velocity

v = Final velocityt =Time interval

a =Acceleration

s =displacement

u , v , a & s can be ‘+’ or ‘-’depend on direction

v = - 4ms-1

Case 1

u = 1ms-1 v = 4ms-1

t = 3s

a = ?v = u + a t 4 = 1 + 3a a = 1ms-2

Case 2‘+’direction

u = 1ms-1t = 5s

v = u + a t - 4 = 1 + 5a a = -1ms-2

a is ‘-‘. Its direction is to the left. a

s

u = 10ms-1A C

B a. What is the time taken bythe stone to travel frompoint A to point B?

u = 10 ms-1, v = 0 ms-1,a = -10 ms-2, t = ?

s 1 10

100

Apply

a

uvt

atuv

s

u = 10ms-1A C

B b. What is the distancetravelled by stone in part(a)? u = 10 ms-1, v = 0 ms-1,t = 1 s, S = ?

m 5 2

)100(

2

)(Apply

tuvS

s

u = 10ms-1A C

B c. What is the time taken by thestone to travel from point B topoint C?

u = 0 ms-1, a = 10 ms-2,S = 5 m, t = ?

s 1

)10(2

1 0 5

2

1 Apply

2

2

t

t

atutS

s

u = 10ms-1A C

B

u = 0 ms-1, t = 1 s,a = 10 ms-2, v = ?

d. What is the velocityof the stone at C?

)(downwards ms 10

)1)(10(0

Apply

1-

atuv

s

u = 10ms-1A C

Be. What is the time taken for the

stone to reach the ground fromC?

The time taken for stonet = 6 – 1 – 1 = 4 s

s

u = 10ms-1A C

B f. What is the height ofthe building?

u = 10 ms-1, a = 10 ms-2,t = 4 s, S = ?

m 120

4)10(2

1 (10)(4)

2

1 Apply

2

2

atutS

s

u = 10ms-1A C

B g. What is the speed of thestone when it strikes theground?

u = 10 ms-1, t = 4 s,a =10 ms-2, v = ?

)(downwards ms50

4010

)4)(10(10

Apply

1-

atuv

Example 2:

1. accelerates uniformly at 4ms-2

from rest for 4 seconds

2. maintains its velocity for 6seconds.

3. brought to rest at uniform deceleration

in another 3 seconds

(a) What is the velocity of the car just afterfirst 4 seconds?

? seconds, 4 ,4ms ,0ms -2-1 vtau

at u v

1-ms 16

)4)(4(

at v

(b) What is the accelerationduring the last 3 seconds?

? seconds, 3 ,0ms ,16ms -1-1 atvu

ion)deccelerat meanssign (negative

ms 3.5 s 3

ms)160( 2-

1-

t

uva

(c) Find the distance traveled by the carduring the first 4 seconds.

u = 0ms-1, a = 4 ms-2, t = 4 seconds, S = ?2

2

1atutS = 0 + 2

1(4)(4)2 = 32 m

(d) Find the distance traveled by the carduring the later 6 seconds.

Total distance traveled by the car = (16)(6) = 96 m

(e) Find the distance traveled by thecar during the last 3 seconds.

u = 16 ms-1, a = -5.3 ms-2, t = 3 seconds, S = ?

2

2

1atutS = (16)(3) +

2

1(-5.3)(3)2 = 24 m

(f) What is the total distance?

Total distance traveled by the car= 32 m + 96 m + 24 m = 152 m

(g) Find the average speed of the car.

1-ms7.1113

152

t

s

car theof speed Average