# LINEAR MOTION - Form 3 Physics Notes

## Introduction

• The study of motion is divided into two areas namely kinematics and dynamics.
• Kinematics deals with the motion aspect only while dynamics deals with the motion and the forces associated with it.
• There are three common types of motion:
• Linear or translational motion.
• Circular or rotational motion.
• Oscillatory or vibrational motion.
• In this topic, we concentrate on linear motion.
• Note that all motion is relative i.e the state of a body; at rest or in motion, is ONLY true with respect to the observer’s position.

## Terms Associated with Linear Motion

• Distance
• is the length of the path covered by a body.
• It only gives the magnitude but no direction i.e it is a scalar quantity.
• Displacement
• is the distance through which a body travels in a specified direction. It is a vector quantity.
Both distance and displacement are measured in metres.
• Speed
• is the distance covered per unit time.
• Speed = distance/time.
• Velocity
• is the rate of change of displacement.
• Velocity = displacement/time.
• It is a vector quantity.
• When the rate of change of displacement is non-uniform, we talk about average velocity;
• Average velocity= total displacement/total time.
Both speed and velocity are expressed in metre per second (m/s).
• Acceleration
• is the rate of change of velocity.
• Thus, Acceleration= change in velocity/time interval = (final velocity v - initial veolicity u)/time.
• Acceleration is measured in metre per square second (m/s2 ).
• If the velocity of a body decreases with time, its acceleration becomes negative.
• A negative acceleration is referred to as deceleration or retardation.

Example 1.1

1. A body covers a distance of 2m in 4seconds, rests for 2seconds and finally covers a distance of 90m in 6seconds. Calculate its average speed.
Solution:
Average speed = total distance/total time = (2m+90m)/(4s+2s+6s)
= 92m/12s = 72/m/s.
2. A body moves 30m due east in 2seconds, then 40m due north in 4seconds. Determine its:
1. Average speed.
Solution:

Average speed= total distance/time = (30m+40m)/(2s+4s)
=70m/6s = 11.67 m/s.
2. Average velocity.
Solution:
Average velocity= total displacement/time =50m/6s
=8.33m/s.
3.  A body is made to change its velocity from 20m/s to 36 m/s in 0.1s. What is the acceleration produced?
a= (v − u)/t =(36m/s − 20m/s)/0.1s
=160 m/s2.
4. A particle moving with a velocity of 200m/s is brought to rest in 0.02s. What is the acceleration of the particle?
a= (v-u)/t =(0m/s − 200m/s)/0.02
= −200/0.02 = −10,000m/s2.

## Motion Graphs.

• There are two categories; displacement-time graphs and velocity time graphs.

### Displacement-time Graphs

• The slope of a displacement-time graph gives the velocity of the body.
• The various displacement-time graphs are as illustrated below:
• Graph A: the body is at rest i.e there is no change in displacement as time changes. The slope of the graph and hence the velocity is zero.
• Graph B: the body moves with a uniform or constant velocity.
• Graph C: the graph becomes steeper with time. The steeper the slope, the higher the velocity. Thus velocity of the body increases with time. The body is therefore accelerating.
• Graph D: the graph becomes less and less steep with time i.e the body has a higher velocity at the beginning and decreases with time. Therefore, the body is said to be decelerating.

### Velocity-time Graphs

• The slope of a velocity-time graph gives the acceleration of the body.
• Note that the area under a velocity-time
graph gives the distance covered by the body.
• The diagram below shows the possible velocity-time graphs:
• Graph A: the velocity remains constant/uniform as time increases. The slope of the graph and hence the acceleration of the body is zero.
• Graph B: the velocity changes uniformly with time. The body moves with a uniform/constant acceleration.
• Graph C: the acceleration is lower where the graph is gentle and higher where the graph is steeper. Hence the acceleration of the body increases with time.
• Graph D: in this case, the graph is steeper at the beginning and becomes gentle with time. Hence the acceleration of the body decreases with time.

## Determination of Velocity and Acceleration

• Two methods are applicable here:
Method 1: Using appropriate instruments e.g a tape measure and a stop watch to measure the displacement of a body and the duration then applying the formula;
Velocity = total displacement/time taken.
• Method 2: Using a ticker-timer.
• It is used to measure velocity of a body specifically over short distances.
• It consists of an electronic vibrator which makes dots on a moving paper tape attached to the object whose velocity is being measured.
• The dots are made at a certain set frequency. For instance, a ticker-timer whose frequency is 50Hz makes dots at intervals of 0.02s.
• The time interval between successive dots is referred to as a tick .
• The spacing between the dots depends on the manner in which the body is moving i.e moving at constant velocity or with increasing velocity or decreasing velocity.
• Generally, the dots are close together when the velocity is low and wide apart when the velocity is high. There are three possible patterns that can be obtained by a ticker-timer as illustrated below:
1. Moving at constant velocity.
The dots are equally or evenly spaced.
2. Moving with increasing velocity (accelerating).
The spacing between the dots is initially small but increases away.
3. Moving with decreasing velocity (decelerating).
The spacing between the dots is initially large but decreases away.

Example 1.2

1.  A paper tape was attached to a moving trolley and allowed to run through a ticker-timer. The figure below shows a section of the tape.

If the frequency of the ticker-timer is 20Hz, determine:
1. The velocity between AB and CD.
Solution
1tick= 1/20 =0.01s
V AB = 15cm/(5ticks×0.01s) =15cm/0.05s
=300cm/s
V CD =30cm/(5ticks×0.01s) =30cm/0.05s
=600cm/s
2. The acceleration of the trolley.
Solution
Note that the velocities calculated in (a) above are average velocities and as such are taken to be the velocities at the midpoints of AB and CD respectively. Hence, the time taken for the change in velocity is the time between the midpoints of AB and CD.
V AB Δt=2ticks × 0.01=0.2s
Therefore, acceleration=(V CD − V AB )/Δt= (600-300)cm/s/0.2s =3000 cms-2.
2. The figure below represents part of a tape pulled through a ticker-timer by a trolley moving down an inclined plane. If the frequency of the ticker-timer is 50Hz, calculate the acceleration of the trolley.

Note that 1tick=1/50 =0.02s.
Initial velocity u =0.5cm/0.02s= 25cms-1
Final velocity v =5.5cm/0.02s= 125cms-1
Hence, acceleration= (v−u)/Δt=(125 − 25)cm/s/0.2s
=200cms-2

## Equations of Linear Motion

• There are three equations governing linear motion. Consider a body moving in a straight line from an initial velocity u to a final velocity v(u, v≠0) within a time t as represented on the graph below:
• The slope of the graph represents the acceleration of the body;
• Acceleration, a=(v−u)/t.
Therefore, v=u+at…………………………………. (i).
This is the first equation of linear motion.
• The area under the graph (area of a trapezium) gives the displacement of the body.
Hence, displacement s= ½(sum of // sides) × perpendicular height between them.
s= ½(u+v)t.
But v=u+at,
Therefore, s=½{u+(u+at)}t
s=½(2u+at)t
Hence, s=ut+½at2 ………………………………. (ii).
This is the second equation of linear motion.
• Also, rearranging equation (i), we have t=(v−u)/a.
substituting this in equation (ii), we obtain;
s=ut+½at2 =u{(v − u)/a}+½a{(v − u)/a}2.
s=u(v − u)/a + a(v − u)2/2a2 = u(v − u)/a + (v − u)2/2a
s= {2u(v − u) + (v-u)2 }/2a = {2uv − 2u2 +v2 +u2 − 2uv}/2a
s= {v2 − u2 }/2a
2as= v2 − u2
Hence, v2 =u2 + 2as ……………………………….. (iii).
This is the third equation of linear motion.
• The three equations hold for any body moving with uniform acceleration.
• Note that for a body which is retarding, the acceleration a is given a negative sign.

Example 1.3

1. A particle travelling in a straight line at 2m/s is uniformly accelerated at 5m/s2 for 8 seconds. Calculate the displacement of the particle.
Solution
s=ut + ½at2 = (2 × 8)+(½ × 5 × 82 )
=176m.
2. An object accelerates uniformly at 3ms-2. It attains a velocity of 4m/s in 5 seconds.
1. What was its initial velocity?
Solution
v=u+at
u= 4 − (3 × 5) = 4−15 = −11m/s.
2. How far does it travel during this period?
Solution
s=ut+½at2 = (4 × 5)+(½ × 3 × 52 )= 57.5m
3. A car travelling at 20m/s decelerates uniformly at 4m/s2. In what time will it come to rest?
Solution
v=u − at, (a is negative since the body is decelerating).
0=20−4t
t=20/4 =5seconds.

## Motions Under the Influence of Gravity

• These include free fall, vertical projection and horizontal projection.
• The three equations of linear motion hold for motions under the influence of gravity.

### Free fall

• A body falling freely in a vacuum starts from an initial velocity zero and accelerates at approximately 9.8ms-2 towards the centre of the earth.
• This is called the acceleration due to gravity g .
• In this case, the air resistance is assumed to be negligible.
• Note that in a vacuum, a feather and a stone released from the same height will take the same amount of time to reach the surface of the earth.
• Therefore, in the three equations of linear motion u=0m/s, s=h and a=g. thus the three equations become:
v=gt, (from v=u+at)
h=½gt2 , (from s=ut+½gt2 )
v2 =2gh, (from v2 =u2 +2as)
• From the above equations:
v= (2gh)½ , where v is the velocity of the body just before it hits the ground.
h=½gt2 =v2/2g, where h is the height through which the body falls.
t=v/g=(2h/g)½ , where t is the time of flight.

Example 1.4

1. A hammer falls from the top of a building 5m high.
1. How long does it take to reach the ground? Take g=10ms-2.
h=½gt2
5=½ × 10t2
t=1½ = 1s
2. With what velocity does it strike the ground?
v= (2gh)½ = (2 × 10× 5)½ = 10 m/s.

### Vertical Projection

• When a body is projected vertically upwards, it decelerates uniformly due to gravity until its velocity reduces to zero at maximum height.
• After attaining the maximum height, the body then falls back with an increasing velocity.
• The body must be given an initial velocity and attains a final velocity of zero at its maximum height.
• Note that the sign of ‘g’ is negative for a vertical projection. This is because the body moves against gravity.
• Hence the three equations of linear motion become:
v=u − gt, (from v = u + at)
h=ut − ½gt2 , (from s = ut +½gt2 )
v2 =u2−2gh, (from v2 =u2 − 2as)
But at maximum height h max , v=0.
• Thus, the three equations reduce to:
1. gt=u,
2. h=ut − ½gt2
3. u2 =2gh.
• From equation (i), the time taken to attain the maximum height is given by;
t=u/g.
• Similarly, the initial velocity u and the maximum height attained by the body h max can be expressed as:
u=gt=(2ghmax)½
And h max =ut − ½gt2 =u2/2g.
• When the body finally falls back to its point of projection, the displacement of the body will be zero. Substituting this in equation (ii), we obtain;
0=ut − ½gt2
Therefore, 0=t(2u − gt)
And t=0 or t=2u/g, where t=0 is the time at the start of the projection and, t is this is the total time of flight i.e for both upward projection and fall back.
• Note that the total time of flight is twice the time taken to attain maximum height.
• Also, the velocity of the body just before hitting its point of projection as it falls back is the same in magnitude but in opposite direction to its initial velocity; v=−u.

Example 1.5

1.  A bullet is shot vertically upwards and rises to a maximum height of 200m. Calculate:
1.  the initial velocity of the bullet,
u=(2gh max )½ = (2 × 10 × 200)½ = 20 m/s
2. the total time of flight.
t=2u/g=2 × 28.28/= 28.28 s
2. An object is released to fall vertically from a height of 20m. At the same time, another object is projected vertically upwards with a velocity of 40m/s.
1. Calculate the time taken before the two objects meet.
Let the time taken to meet be t. then, after a time t the distance covered by the object moving downwards will be;
sd =½gt2 , (since u=0).
=½ × 10t2 = 5t2
The distance covered by the object projected upwards after a time t will be;
su =ut − ½gt2 = 40t − 5t2
But sd + su =20m
Therefore, 5t2 + 40t − 5t2 =20
t=20/40=0.5s
2. At what height above the point of projection do they meet?
su =ut − ½gt2 =(40×0.5) − (½ × 10×0.52 )
=19.75m

### Horizontal projection

• If two objects A and B at a point some height above the ground are such that A is allowed to fall freely (vertically downwards) while is B given a horizontal projection with an initial velocity u, then both objects take the same duration to reach the ground.
• This is because both are acted on by the same gravitational force. The object on the horizontal projection moves with a constant velocity u. hence, the horizontal acceleration of the object is zero.
• For the object falling freely, the acceleration is equivalent to ‘g’ and the initial velocity u is zero.
• However, the object under horizontal projection will strike the ground some distance away from the point the other object strikes the ground.
• This horizontal distance covered by the object is referred to as the ‘ range R ’.
• Note that both A and B will strike the ground with the same velocity.
• Since a=0 for the horizontal projection, s=R=ut.
• Also, the time taken to reach the ground in both cases is expressed as;
t = u/g.

Example 1.6

1. A stone is thrown at a velocity of 30m/s to the horizontal by a girl at the top of a tree whose height is 30m.
Calculate:
1. the time taken for the stone to strike the ground.
Since both free fall and horizontal projection take the same duration;
h=½gt2
30=½ × 10 × t2
t=6½ = 2.45 s
2. the velocity at which the stone strikes the ground.
Therefore, v=(2gh)½ =(2 × 10 × 30)½
=24.495 m/s
2. A jet fighter on practice moving at a velocity of 20m/s released a bomb above the ground which hits the ground after 3s. Calculate:
1. the distance from the ground to the jet,
h=½gt2 =½ × 10 × 32
=45m
2. the horizontal distance from the target when the bomb is released.
R=ut=20 × 3
=60m.

### Experimental Determination of Acceleration due to Gravity.

This can be done as follows:

• Set the apparatus as shown in the diagram above.
• Set the length of the string at 30cm. note that the length l is measured from the centre of the bob.
• Displace the bob sideways through a small angle of about 2 0 and release it so as to oscillate.
• With the help of a stop watch, measure and record the time for ten oscillations (allow some little oscillations after release before timing). Repeat this step twice or thrice and determine the average time.
• Hence calculate the period T(time for one oscillation).
• Repeat the above steps for l=40cm, 50cm, 60cm, 70cm and 80cm. complete the table below:
 Length Time for oscillations, t(s) Period, T(s) T2(s2) t1 t2 t3 t=t1+t2+t3/3 30 40 50 60 70 80

#### Observations and Conclusion

• The frequency of oscillation increases with decrease in length of the string. A graph of T 2 against length l is a straight line through the origin.
• Generally, a graph of T 2 against length for a simple pendulum satisfies the equation T 2 =4π 2 l/g.
• Hence, the slope of the graph above is equals to 4π2 /g.

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