Chapter 2:Motion in One DimensionEXAMPLES

Example 2.1 Displacement

x1 = 30 m x2 = 10 m Displacement is a VECTOR

10 30 20f ix x x m m m

Example 2.2 Average Velocity & Speed

Suppose the person walks during 50 seconds.

Displacement

Distance (d) = 100m Average velocity:

Average Speed:

t

xx

t

xvv if

xaverage

XfXi

40 0 40f ix x x m m m

400.8 /

50average x

x mv v m s

t s

total distance 100Average speed 2.0 /

total time 50

mm s

s

If an objects moves at uniform velocity (constant), then: Instantaneous velocity and average velocity at any instant (t) are the same.

Instantaneous = average

Example 2.3 Instantaneous & average velocities

Are Instantaneous velocity and Average velocity at any instant t the same? NOT ALWAYS!!!!

Example: A car starts from rest, speed up to 50km/h, remains at that speed for a time. Slow down to 20 km/hr in a traffic jam, the finally stops. Traveling a total of 15 km in 30 min (0.5 hr).

Example 2.4 Instantaneous & average velocities

Example 2.4, cont

15Average velocity 30.0 /

0.50

x kmkm h

t h

2.4 Acceleration

Example 2.5 Average Acceleration

ax (+), vx(+) Speeding Up!!

ax (), vx() Speeding Up!!

2/2.43600

10001515

00.5

0/75sm

ss

m

sh

km

s

hkm

tt

vva

if

ifx

2/0.20.5

/0.10

00.5

)/5(/0.15sm

s

sm

s

smsm

tt

vva

if

ifx

Example 2.6 Average Acceleration

ax (+), vx() Slowing Down!!

ax (), vx(+) Slowing Down!!

2/0.20.5

/0.10

00.5

)/0.15(/0.5sm

s

sm

s

smsm

tt

vva

if

ifx

2/0.20.5

/0.10

00.5

/0.15/0.5sm

s

sm

s

smsm

tt

vva

if

ifx

Example 2.7 Conceptual Question

Velocity and acceleration are both vectors (they have magnitude & direction).

Are the velocity and the acceleration always in the same direction?

NO WAY!!

Example 2.8 Conceptual Question

Velocity and acceleration are both vectors (they have magnitude & direction).

Is it possible for an object to have a zero acceleration and a non-zero velocity?

YES!!! Drive 65 miles/h on the Freeway

Example 2.0 Conceptual Question

Velocity and acceleration are both vectors (they have magnitude & direction).

Is it possible for an object to have a zero velocity and a non-zero acceleration?

YES!!!Start your car!!!

Examples to Read!!! Example 2.5 (Text book Page 31) Example 2.8 (Text book Page 37)

Material for the Midterm

2.6 Constant Acceleration

Initial velocity at A is upward (+) and acceleration is g (– 9.8 m/s2)

At B, the velocity is 0 and the acceleration is g (– 9.8 m/s2)

At C, the velocity has the same magnitude as at A, but is in the opposite direction

The displacement is – 50.0 m (it ends up 50.0 m below its starting point)

Example 2.10 Free Fall Example

(1) From (A) → (B)

Vyf(B) = vyi(A) + ayt(B)

0 = 20m/s + (–9.8m/s2)t(B)

t = t (B) = 20/9.8 s = 2.04 s

ymax = y(B) = y(A) + vyi(A)t + ½ayt2

y(B) = 0 + (20m/s)(2.04s) + ½(–9.8m/s2)(2.04s)2

y(B) = 20.4 m

Example 2.10, cont

Example 2.10, cont

(2) From (B) → (C): y(C) = 0

y(C) = y(A)+ vyi(A) t – ½ayt2

0 = 0 + 20.0 t – 4.90t2

(Solving for t): t(20 – 4.9t) = 0 t = 0 or t(C) = t = 4.08 s

vyf(C) = vyi(A) + ayt (C)

vyf(C) = 20m/s + (– 9.8m/s2)(4.08 s)

vyf(C) = –20.0 m/s

(3) From (C) → (D)Using position (C) as the reference point the t at (D) position is not 5.00s.

It will be:

t (D) = 5.00 s – 4.08 s = 0.96 vyf(D) = vyi(C) + ayt (D) vyf(D) = -20m/s + (– 9.8m/s2)(0.96 s)

vyf(D) = – 29.0 m/s

y(D) = y(C) + vyi(C)t + ½ayt2

y(D) = 0 – (29.0m/s)(0.96s) – (4.90m/s2)

(0.96s)2 = – 22.5 m y(D) = –22.5 m

Example 2.10, cont

Example 2.11 (Problem #66 page 54)

From the free fall of the rock the distance will be: From the sound de same distance will be: But: t1 + t2 = 2.40s t1 = 2.40 – t2

Replacing (t1 ) into the first equation and equating to the second:

.

21

19.80

2d t

2336d t

22 2336 4.90 2.40t t

22 24.90 359.5 28.22 0t t

2

2

359.5 359.5 4 4.90 28.22

9.80t

2

359.5 358.750.0765 s

9.80t

2336 26.4 md t

Example 2.12 Objective Question #13 A student at top of the building of height h throws one ball upward

with speed vi and then throws a second ball downward with the same initial speed, vi . How do the final velocities of the balls compare when they reach the ground?

After Ball 1 reaches maximum

height it falls back downward

passing the student with velocity –vi . This

velocity is the same as Ball 2 initial velocity, so

after they fall through equal height h,

their impact speeds will also be the same!!!

+vi

- vi

BALL 1

h hh

BALL 2

- vi

Material from the book to Study!!! Objective Questions: 2-13-16 Conceptual Questions: 6-7-9 Problems: 3-6-11-16-17-20-29-42-44

Material for the Midterm