## Introduction

Study of motion is divided into two;

- Kinematics
- Dynamics

In kinematics forces causing motion are disregarded while dynamics deals with motion of objects and the forces causing them.

- Displacement

Distance moved by a body in a specified direction is called displacement. It is denoted by letterâ€˜sâ€™ and has both magnitude and direction. Distance is the movement from one point to another. The Si unit for displacement is the metre (m).

- Speed

This is the distance covered per unit time.

Speed= distance covered/ time taken. Distance is a scalar quantity since it has magnitude only.

The SI unit for speed is metres per second(m/s or ms^{-1})

Average speed= total distance covered/total time taken Other units for speed used are Km/h.

Examples

- A body covers a distance of 10m in 4 seconds. It rests for 10 seconds and finally covers a distance of 90m in 60 seconds. Calculate the average speed.

Solution

Total distance coveredÂ =10+90= 100m

Total time takenÂ =4+10+6= 20 seconds

Therefore average speedÂ = 100/20= 5m/s

- Calculate the distance in metres covered by a body moving with a uniform speed of 180 km/h in 30 seconds.

Solution

Distance covered=speed*time

=180*1000/60*60=50m/s

=50*30

=1,500m

- Calculate the time in seconds taken a by body moving with a uniform speed of 360km/h to cover a distance of 3,000 km?

Solution

Speed:360km/h=360*1000/60*60=100m/s

Time=distance/speed

3000*1000/100

=30,000 seconds.

## III. Velocity

This is the change of displacement per unit time. It is a vector quantity.

Velocity=change in displacement/total time taken

The SI units for velocity are m/s

Examples

- A man runs 800m due North in 100 seconds, followed by 400m due South in 80 seconds. Calculate,
- His average speed
- His average velocity
- His change in velocity for the whole journey

Solution

- Average speed: total distance travelled/total time taken

=800+400/100+80

=1200/180

=6.67m/s

- Average velocity: total displacement/total time

=800-400/180

=400/180

=2.22 m/s due North

Change in velocity=final-initial velocity

= (800/100)-(400-80)

=8-5

=3m/s due North

- A tennis ball hits a vertical wall at a velocity of 10m/s and bounces off at the same velocity. Determine the change in velocity.

Solution

Initial velocity(u)=-10m/s

Final velocity (v) = 10m/s

Therefore change in velocity= v-u

=10- (-10)

=20m/s

## Acceleration

This is the change of velocity per unit time. It is a vector quantity symbolized by â€˜aâ€™. Acceleration â€˜aâ€™=change in velocity/time taken= v-u/t The SI units for acceleration are m/s2 Examples 1. The velocity of a body increases from 72 km/h to 144 km/h in 10 seconds.

Calculate its acceleration.

Solution

Initial velocity= 72 km/h=20m/s

Final velocity= 144 km/h=40m/s

Therefore â€˜aâ€™ =v-u/t

= 40-20/10

2m/s2

- A car is brought to rest from 180km/h in 20 seconds. What is its retardation?

Solution

Initial velocity=180km/h=50m/s

Final velocity= 0 m/s

A = v-u/t=0-50/20

= -2.5 m/s2

Hence retardation is 2.5 m/s2

## Motion graphs

Distance-time graphs

Area under velocity-time graph

Consider a body with uniform or constant acceleration for timeâ€˜tâ€™ seconds;

This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body underâ€˜tâ€™ seconds.

Example

A car starts from rest and attains a velocity of 72km/h in 10 seconds.

It travels at this velocity for 5 seconds and then decelerates to stop after another 6 seconds.

Draw a velocity-time graph for this motion. From the graph;

- Calculate the total distance moved by the car
- Find the acceleration of the car at each stage.

Solution

- From the graph, total distance covered= area of (A+B+C)

=(1/2Ã—10Ã—20)+(1/2Ã—6Ã—20)+(5Ã—20)

=100+60+100

=260m

Also the area of the trapezium gives the same result.

- Acceleration= gradient of the graph

Stage A gradient= 20-0/ 10-0 = 2 m/s2

Stage b gradient= 20-20/15-10 =0 m/s2

Stage c gradient= 0-20/21-15 =-3.33 m/s2

## Using a ticker-timer to measure speed, velocity and acceleration.

It will be noted that the dots pulled at different velocities will be as follows;

Most ticker-timers operate at a frequency of 50Hzi.e. 50 cycles per second hence they make 50 dots per second. Time interval between two consecutive dots is given as,

1/50 seconds= 0.02 seconds. This time is called a tick.

The distance is measured in ten-tick intervals hence time becomes 10Ã—0.02= 0.2 seconds.

Examples

- A tape is pulled steadily through a ticker-timer of frequency 50 Hz.

Given the outcome below, calculate the velocity with which the tape is pulled.

Solution

Distance between two consecutive dots= 5cm

Frequency of the ticker-timer=50Hz

Time taken between two consecutive dots=1/50=0.02 seconds

Therefore, velocity of tape=5/0.02= 250 cm/s

- The tape below was produced by a ticker-timer with a frequency of 100Hz. Find the acceleration of the object which was pulling the tape.

Solution

Time between successive dots=1/100=0.01 seconds

Initial velocity (u) 0.5/0.01 50 cm/s

Final velocity (v) 2.5/0.01= 250 cm/s

Time taken= 4 Ã—0.01 = 0.04 seconds

Therefore, acceleration= v-u/t= 250-50/0.04=5,000 cm/s2

Equations of linear motion

The following equations are applied for uniformly accelerated motion;

v = u + at

s = ut + Â½ at2

v2= u2 +2as

Examples

- A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.

Solution

V2 = u2 +2as

= (60) +2Ã—10Ã—320

=3600+6400

= 10,000

Therefore v= (10,000)1/2

v= 100m/s

A body whose initial velocity is 30 m/s moves with a constant retardation of 3m/s. Calculate the time taken for the body to come to rest.

Solution

v = u+at

0= 30-3t

30=3t

t= 30 seconds.

- A body is uniformly accelerated from rest to a final velocityof 100m/s in 10 seconds. Calculate the distance covered.

Solution

s=ut+ Â½ at2

=0Ã—10+ Â½ Ã—10Ã—102

= 1000/2=500m

## Motion under gravity.

- Free fall

The equations used for constant acceleration can be used to become,

v =u+gt

s =ut + Â½ gt^{2}

v2= u+2gs

- Vertical projection

Since the body goes against force of gravity then the following equations hold

v =u-gt â€¦â€¦â€¦â€¦â€¦1

s =ut- Â½ gt2 â€¦â€¦2

v2= u-2gs â€¦â€¦â€¦â€¦3

N.B time taken to reach maximum height is given by the following

t=u/g since v=0 (using equation 1)

## Time of flight

The time taken by the projectile is the timetaken to fall back to its point ofprojection. Using eq. 2 then, displacement =0

0= ut- Â½ gt^{2}

0=2ut-gt2

t(2u-gt)=0

Hence, t=0 or t= 2u/g

t=o corresponds to the start of projection

t=2u/gcorresponds to the time of flight

The time of flight is twice the time taken to attain maximum height.

Maximum height reached.

Using equation 3 maximum height, Hmax is attained when v=0 (final velocity).

Hence v2= u2-2gs;- 0=u2-2gHmax, therefore

2gHmax=u2

Hmax=u2/2g

## Velocity to return to point of projection.

At the instance of returning to the original point, total displacement equals to zero.

v2 =u2-2gs hence v2= u2

Thereforev=u or v=Â±u

Example

A stone is projected vertically upwards with a velocity of 30m/s from the ground.

Calculate,

- The time it takes to attain maximum height
- The time of flight
- The maximum height reached
- The velocity with which it lands on the ground. (take g=10m/s)

Solution

- Time taken to attain maximum height

T=u/g=30/10=3 seconds

- The time of flight

T=2t= 2Ã—3=6 seconds

Or T=2u/g=2Ã—30/10=6 seconds.

- Maximum height reached

Hmax= u2/2g= 30Ã—30/2Ã—10= 45m

- Velocity of landing (return)

v2= u2-2gs, but s=0,

Hence v2=u2

Thereforev=(30Ã—30)1/2=30m/s

## Horizontal projection

The path followed by a body (projectile) is called trajectory. The maximum horizontal distance covered by the projectile is called range. The horizontal displacement â€˜Râ€™ at a timeâ€˜tâ€™ is given by s=ut+1/2 at 2 Taking u=u and a=0 hence R=ut, is the horizontal displacement and h=1/2gt2 is the vertical displacement.

NOTE

The time of flight is the same as the time of free fall.

Example

A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s.

Calculate,

- The time taken by the ball to strike the ground
- The distance from the foot of the cliff to where the ball strikes the ground.
- The vertical velocity at the time it strikes the ground. (take g=10m/s)

Solution

- h= Â½ gt2

20= Â½ Ã—10Ã—t2

40=10t2

t2=40/10=4

t=2 seconds

- R=ut

=10Ã—2

=20m

- v=u+at=gt

= 2Ã—10=20m/s