CHAPTER 2

MOTION IN ONE DIMENSION (KINEMATICS)

• Why objects move

• Displacement and velocity • Average velocity and average speed • Relative velocity • Instantaneous velocity • Displacement-time graphs

• Acceleration • Average acceleration • Instantaneous acceleration • Velocity-time graphs • Equations of motion

• Free fall

This chapter is about MOTION ... so, why do objects move?

It is FORCE that causes a change in motion (starting, stopping, going round corners, etc).

It is the force of the road on the tires that cause a car to move!

It is the force of gravity downward that makes the cat fall! (Its weight.)

It is the force exerted by the typists fingers that makes the keys move!

For the moment we won’t worry about what causes motion but the rules that tell us how objects move.

Displacement and distance travelled

Displacement is . Distance travelled is also.

Displacement is . Distance travelled is .

Displacement is . Distance travelled is .

Note: • distance travelled displacement, • displacement can be positive or negative, • distance always .

start finish

�

(x2 − x1) = Δx

(x2 − x1)

start finish

(x3 − x1)

(x3 − x1)

(x3 − x1) = (d1 + d2)

(d1 + d2)

start finish

�

≥

≥ 0

Average velocity and average speed

• Average velocity .

• Average speed .

In this example, ,

i.e., average speed > average velocity

Average speed always Average velocity can be

Dimension: velocity and speed

Units: .

start finish

Δt = t2 − t1

vav = displacement

time interval= Δx

Δt

= distance

time interval=

d1 + d2Δt

(d1 + d2) > (x3 − x1)

⇒ [L]

[T]

m/s, km/h, ft/s, mi/h, etc.

> 0. ≤ 0.

Question 2.1: A runner runs for 2.5 km on a straight path

in 9.0 mins and takes 30.0 mins to walk back to the

starting point.

(a)  What is the runner’s velocity for

(b) What is the average velocity for the time spent walking?

(c) What is the average speed for the whole trip?

(d) What is the average velocity for the whole trip?

(e) Draw a position-time graph for the problem.

0 > t ≥ 9 mins? vav =

(x2 − x1)Δt1

= 2.50 km9 min

(a)  From :

(b) From :

(c) Average speed

(d)

x1 → x2

= 2500 m

9 × 60 s= 4.63 m/s.

x2 → x1 vav =

(x1 − x2)Δt2

= −2.50 km30 min

= −2500 m

30 × 60 s= −1.39 m/s.

= total distance

total time= 5000 m

39 × 60 s= 2.14 m/s.

vav = displacement

total time= 0.

(e) The position-time graph for the problem. The runner’s

displacement is the position relative to the starting point.

• The slope of AB is , i.e., the

average velocity for the first 9 mins.

• The slope of BC is , i.e., the

average velocity while walking.

Important concept: the average velocity is the slope of the

displacement-time graph.

2.50 km9 min

⇒ 4.63 m/s

−2.50 km

30 min⇒−1.39 m/s

Relative velocity

If the velocity of the walkway relative to the ground (i.e.,

the stationary observer) is vwg, and the velocity of 2

with respect to 1 is v21, then the velocity of 2 with

respect to a stationary observer (the relative velocity) is:

(vwg + v21)

Note, if 2 is walking in the opposite direction, his

velocity relative to the observer is

(vwg − v21)

In that case, what “happens” if vwg = v21 ?

This is called a Galilean transformation.

vwg

v21 1 2

Question 2.2: Two trains, 35 km apart and traveling

at 10 km/h, are approaching each other on parallel

tracks. A bird flies back and forth between the trains

at 15 km/h relative to the ground until the trains pass

each other.

(a) How long is the bird flying?

(b) What distance does the bird fly?

35 km

10 km/h 10 km/h

Assume the bird flies a total distance d.

Av. speed of bird is

where t is the elapse time before the trains meet.

Imagine you are sitting on the front of the left hand

train, then the relative speed of approach of the right

hand train to you is

So the time taken for the trains to pass is

distance flown

time= d

t= 15 km/h,

i.e., d = (15 km/h) × t,

(10 km/h) + (10 km/h) = 20 km/h.

t = distance they travel

relative speed= 35 km

20 km/h= 1.75 h.

∴d = (15 km/h)(1.75 h) = 26.3 km.

35 km

10 km/h 10 km/h

Question 2.3: A river flows from left to right with a

velocity of 0.20 m/s relative to the bank. If you swim

with a velocity of 1.00 m/s relative to the water, does it

take you more time, less time, or the same time to swim

from than if the river was not flowing at all? A → B→ A

When you swim from your velocity relative to

the bank is

When you swim from your velocity relative to

the bank is

If the distance AB is d meters, then the times taken (in

seconds) are:

So the total elapsed time is

If the river is not flowing then

i.e., it takes more time if the river is flowing!

A → B

vAB = 1.00 m/s + 0.20 m/s = 1.20 m/s.

B→ A

vBA = 1.00 m/s − 0.20 m/s = 0.80 m/s.

tAB = d

1.20 m/s and tBA = d

0.80 m/s

tAB + tBA = (1.20 m/s + 0.80 m/s)d

(0.80 m/s)(1.20 m/s)= 2.08d s.

′t = d

1.00 m/s+ d

1.00 m/s= 2.00d s,

Instantaneous velocity

Velocities are rarely constant. Consider the displacement

time graph below in which the velocity changes with time.

The average velocity

from is

Note that

We define the instantaneous velocity at the point as the

slope of the tangent at the point

i.e., the first derivative of x with respect to t at the point .

P ′2 P1 → P2

v1→2 = Δx

Δt=

x2 − x1t2 − t1

.

P1

P1

∴v(t) = LimitΔt→0ΔxΔt

= dxdt

⎡

⎣⎢

⎤

⎦⎥

P1

,

P1,

v1→2 < v1→ ′2 .Question 2.4: The plot shows the variation of the

position of a car with time.

(a) What was its velocity at ?

(b) During the time interval was the car

speeding up, slowing down or moving at constant

speed?

t = 15 s

0 < t <10 s

(a)  Velocity is defined as the slope at

i.e.,

t = 15 s,

v = dx

dt⎛⎝⎜

⎞⎠⎟ t=15 s

= Δx

Δt= (2 m − 4 m)

(20 s −10 s)= −0.20 m/s.

(b) The instantaneous speed is given by the magnitude

of the slope of the displacement-time curve.

The magnitude of the slope increases over the

time interval . Therefore, the car was

speeding up.

Note, in the interval the speed was

constant. During the interval the car was

slowing down.

v = dx

dt 0 < t ≤10 s

10 s ≤ t ≤ 20 s

20 s ≤ t ≤ 30 s

Whenever velocity changes we have … acceleration.

Average acceleration

Dimension:

Units:

Instantaneous acceleration

Since then also.

aav = Δv

Δt=

v2 − v1t2 − t1

.

[L][T]

1[T]

= [L]

[T]2⇒ [L][T]−2.

m/s2, ft/s2, etc.

a = LimitΔt→0

ΔvΔt

= dvdt

.

v = dx

dt,

a = d2x

dt2

Equations of motion

To start with, assume that the acceleration a is constant.

Since we can integrate this expression with

respect to t and obtain:

where is an integration constant. If the initial

velocity (when ) is , then i.e.,

Also, since

where is another constant. If when

then so

i.e., the displacement is

dvdt

= a,

v = dv = a∫∫ dt = at +C1,

C1

t = 0 v! C1 = v!,

v = v! + at ... ... ... ... (1)

v = dx

dt,

x = dx∫ = vdt = (v! + at)dt∫∫

= v!t +

12

at2 +C2,

C2 x = x! t = 0,

C2 = x!,

x = x! + v!t +

12

at2,

(x − x!) = v!t +

12

at2 ... ... ... (2)

Since the acceleration is constant we can define the

average velocity

Then the displacement is

But, from (1)

and substituting for t in (4) we find

Equations (1) – (5) are the principle kinematic equations

for one-dimensional motion at constant acceleration.

vav = 1

2(v + v!) ... ... ... ... (3)

Δx = x − x! = vavt = 1

2(v + v!)t ... (4)

t =

(v − v!)a

,

x − x! =

12

(v + v!)(v − v!)

a

=

(v2 − v!2)

2a.

∴v2 = v!2 + 2a(x − x!) ... ... ... (5)

If the acceleration is not constant, providing we know how

the velocity varies with time, we can still determine the

displacement over a time interval. For example, given

the velocity-time variation shown above, what is the

displacement of an object over the time period

We split the time period into a

large number of equal time

intervals, If is very small,

the displacement from to

is so the total displacement from is

t1 → t2 ?

Δt. Δt

t t + Δt

Δx ≈ viΔt, t1 → t2

x(t2 ) − x(t1) ≈ v1Δt + v2Δt +! ≈ (viΔti∑ ).

The sum becomes exact if Then

i.e., the displacement is the area under the velocity-time

plot from to So, if we know how the velocity

varies with time, we can determine the

displacement over any time interval.

Note, the above expression is valid even if the

acceleration is changing, when our previous equations

of motion (1) – (5) cannot be applied.

Δt → 0.

x(t2 ) − x(t1) = LimitΔt→0 viΔti∑( ) = vdt,

t1

t2

∫

t1 t2.

i.e., v → v(t),

Question 2.5: A car starts from rest at and

accelerates at a constant rate of for 5.0 s. It

then travels at a constant speed for 15.0 s, when the

driver is forced to slam on the brakes. The car comes to

stop after a distance of 8.0 m.

(a) What is the velocity of the car after the initial 5.0 s?

(b) After the driver slams on the brakes and the car

comes to a stop, what is the acceleration?

(c) When the car comes to stop, how far is it from the

starting point?

x! = 50.0 m

4.0 m/s2

t = 0 t1 t2 t3

The accelerations from are

constant, so we can use eqs. (1) – (5).

(a)  Using (1) we have

(b)  To find the acceleration we use eq. (5):

where d is distance traveled from .

The acceleration is negative because the car is slowing.

(c) Here are two ways to find the total distance.

• Find the distance traveled over each time interval.

• Plot the velocity-time graph and find the area.

v = v! + at = 0 + (4.0 m/s2)(5.0 s) = 20.0 m/s.

t2 → t3

v2 = v!

2 + 2ad ⇒ a =v2 − v!

2

2d

∴ a = 0 − (20.0 m/s)2

2(8.0 m)= −25.0 m/s2.

t = 0 to t1 and from t2 to t3

From and using eq. (2) we have

Since the velocity is constant from , the distance

traveled is

From the distance traveled is 8.0 m, so the total

distance traveled by the car is

To confirm this result, we plot the velocity-time graph,

but we need , i.e., the time for the car to stop.

Using eq. (1) we find

Here is the velocity-time graph …

t → t1

x − x! = v!t +

12

at2 = 0 + 12

(4.0 m/s2)(5.0 s)2 = 50.0 m.

t1 → t2

v(t2 − t1) = (20.0 m/s)(15.0 s) = 300 m.

t2 → t3

(50.0 m) + (300 m) + (8.0 m) = 358 m.

(t3 − t2 )

(t3 − t2 ) =

v − v!a

= 0 − 20.0 m/s

−25.0 m/s2= 0.8 s.

time (s)

velocity

20.0 (m/s)

0 5 20 20.8

A1 A2 A3

The total distance traveled is given by the area under

the velocity-time graph, i.e., A1 + A2 + A3

= 12

(5.0 s)(20.0 m/s) + (15.0 s)(20.0 m/s)

+ 12

(0.8 s)(20.0 m/s)

= 358 m.

Question 2.6: An object, initially at rest at

experiences an acceleration given by

After any time t, what is

(a)  the instantaneous velocity, and

(b)  the position of the object relative to the origin?

(c)  What is the displacement of the object over the

time period

x! = 1.00 m,

a(t) = 1.50 + 0.20t m/s2.

2.0 s ≥ t ≥ 4.0 s?

Because the acceleration is not constant we cannot use eqs.

(1) – (5); we must use calculus.

(a)  Given then

Since the initial velocity is zero at then .

(b) Since the position is given by

The initial position at is then

a(t) = dv

dt= 1.50 + 0.20t,

v = dv = (1.50 + 0.20t)dt∫∫

= 1.50t + 0.20 12

t2⎛⎝⎜

⎞⎠⎟+C1.

t = 0, C1 = 0.

∴v(t) = 1.50t + 0.10t2 m/s.

v = dx

dt,

x = dx∫ = vdt =∫ 1.50t + 0.10t2( )dt∫

= 1.5012

t2⎛⎝⎜

⎞⎠⎟+ 0.10

13

t3⎛⎝⎜

⎞⎠⎟+C2.

t = 0 x! = 1.00 m, C2 = 1.00 m.

∴x(t) = 1.00 + 0.75t2 + 0.033t3 m.

(c) By definition, the displacement is

Using the result in part (b)

Check using (b):

Δxt1→t2

= vdt =t1

t2∫ 1.50t + 0.10t2( )

2

4∫ dt.

1.50t + 0.10t2( )2

4∫ dt = 0.75t2 + 0.033t3 +C2

⎡⎣⎢

⎤⎦⎥2

4

= 0.75 42 − 22( ) + 0.033 43 − 23( )= 10.85 m.

x(t = 2) = 1.00 + 0.75(22) + 0.033(23) = 4.26 m.

x(t = 4) = 1.00 + 0.75(42) + 0.033(43) = 15.11 m.

∴Δx = (15.11− 4.26)m = 10.85 m.

Question 2.7: An object, initially at rest at the origin,

experiences an acceleration

(a) What is its speed when

(b) How far from the origin is it at that time?

a(t) = 1.50e−0.3t m/s2.

t = 5 s?

(a)

But

(b)

We have a(t) = dv(t)dt

= 1.50e−0.3t m/s2.

∴v(t) = dv(t)∫ = 1.50e−0.3t∫ dt

= − 1.50.3

e−0.3t +C1 = −5e−0.3t +C1.

We have v(t) = dx(t)dt

= 5 1− e−0.3t( ) m/s.

∴x(t) = dx(t) = 5 1− e−0.3t( )∫∫ dt = 5 t + e−0.3t

0.3

⎛

⎝⎜

⎞

⎠⎟ +C2.

But x(0) = 0 so C2 = −16.67.

∴x(t) = 5 t + e−0.3t

0.3

⎛

⎝⎜

⎞

⎠⎟ −16.67 m.

When t = 5 s, x(5s) = 5 5+ e−1.5

0.3

⎛

⎝⎜

⎞

⎠⎟ −16.67

= 12.05 m.

v(0) = 0 so C1 = 5. ∴v(t) = 5 1− e−0.3t( ) m/s.

When t = 5 s, v(5 s) = 5 1− e−1.5( )= 5(1− 0.223) = 3.88 m/s.

Free fall – Galileo Galilei (1564-1642)

In “free fall”, i.e., in the absence of air

resistance, all objects close to the

Earth’s surface fall with a constant

acceleration (directed downward)

known as the “acceleration due to

gravity”.

Therefore, the velocity of the object in free fall increases

by 9.81 m/s each second. For an object dropped from

rest at a height its velocity and vertical displacement

are

and

Note that neither the velocity nor the displacement

depend on the mass of the object!

a ⇒ g = 9.81 m/s2 (32.2 ft/s2 )

y! : t = 0

y : t

y!,

y − y! = v!t +

12

at2 = 12

gt2.

v = v! + at = gt

So, in the absence of air resistance, all objects fall at the

same rate:

and all objects fall the same distance in the same time:

The velocity of an object in free-fall, starting from rest.

velo

city

(m/s

) D

ista

nce

(m) The distance fallen by

an object in free-fall, starting from rest.

Time elapse photographs of an

apple and feather released at

the same time, in the absence of

air resistance (free fall).

Apollo 15 astronaut David Scott drops a geological

hammer (1.32 kg) and a falcon feather (0.03 kg) at the

same time on the Moon (July/August 1971).

Question 2.8: A ball is dropped from rest from the top of

a tower 20.0 m high. A second ball is thrown vertically

upwards from the ground at the same instant the first ball

is released. If the balls pass each other at the half-height

point, i.e., a height of 10.0 m above the ground,

(a) how long does each ball take to reach the half-

height point?

(b) What is the initial speed of the second ball?

(a)  Since the balls are released at the same time, when

they pass, they will have traveled for the same time

interval. We take the upward direction as the positive

direction. So, for the first ball the mid way point is at

To find the time to reach the midway point we

use

i.e.,

(b) The time for the second ball to reach the half-height

point is the same but the displacement of the second ball

is upwards. For the second ball,

(y − y!) = v!t +

12

at2, t =

2(y − y!)a

y = 10 m.

= 2(10 m − 20 m)

−9.81 m/s2= 1.43 s.

(y − y!) = v!t +

12

at2 = v!t −12

gt2.

∴v! =

(y − y!)t

+ 12

gt = 10 m1.43 s

+ 12

(9.81 m/s2)(1.43 s)

= 14.0 m/s.

Question 2.9: A rocket is fired vertically with a constant

upward acceleration of After 25.0 s, the engine

is shut off but the rocket continues to rise until it reaches

its maximum height. It then falls back to the ground.

Neglecting air resistance, find

(a) the height the rocket reaches,

(b) the time to reach maximum height,

(c) the velocity of the rocket just before it strikes the

ground.

20.0 m/s2.

Take the upward direction as positive.

(a)

i.e.,

Also

Between and (free fall).

i.e.,

Maximum height:

(y1 − y!) = v!(t1 − t!) +

12

a(t1 − t!)2,

y1 =

12

(20.0 m/s2)(25.0 s)2 = 6250 m.

v1 = v! + at = (20.0 m/s2)(25.0 s) = 500 m/s.

y1 y2, a = −g = −9.81 m/s2

v22 = v1

2 + 2a(y2 − y1),

(y2 − y1) =

v22 − v1

2

2a= 0 − (500 m/s)2

2(−9.81 m/s2)= 12700 m.

y2 = 6250 m +12700 m = 18950 m.

y! = 0 : v! = 0 : t! = 0

y1 : v1 : t1 = 25.0 s

y2 : v2 = 0 : t2

a = 20.0 m/s2

Free fall: a = −g

y3 = 0 : v3

(b) Total climb time is But

i.e.,

(c) i.e.,

The appropriate solution is , as we took

the upward direction as positive but the velocity is

directed downward.

t1 + t2.

v2 = v1 + a(t2 − t1),

(t2 − t1) =

v2 − v1a

= 0 − (500 m/s)

−9.81 m/s2= 51.0 s.

∴ t2 = 25.0 s + 51.0 s = 76.0 s.

v32 = v2

2 + 2a(y3 − y2) v32 = 2(−g)(−y2) = 2gy2

= 2(9.81 m/s2)(18950 m) = 3.72 ×105 (m/s)2.

∴v3 = ±610.2 m/s.

v3 = −610.2 m/s

y! = 0 : v! = 0 : t! = 0

y1 = 6250 m : v1 = 500 m/s : t1 = 25.0 s

y2 = 18950 m : v2 = 0 : t2

a = 20.0 m/s2

Free fall: a = −g

y3 = 0 : v3

Question 2.8: If, in the photograph, the upper ball is at the top of its trajectory, estimate the time it takes for a ball to loop from one hand to the other.

DISCUSSION PROBLEM [2.1]:

In the photograph, the upper ball is at the top of its

trajectory. Estimate the time it takes for a ball to loop

from one hand to the other. It is about

A 0.5 s B 0.75 s C 1.0 s D 1.25 s?

Galileo investigated the acceleration of gravity by rolling

balls down a smooth incline. So, how is the acceleration

due to gravity affected by the slope of a frictionless

incline?

On a vertical surface the acceleration is the same as in free fall, i.e.,

(θ = 90!),

a = g.

On a horizontal surface gravity does not produce an acceleration along the surface, i.e.,

(θ = 0),

a = 0.

On a surface inclined at an angle to the horizontal, the acceleration is

θ

a = gsinθ.