Linear Motion - High School Physics Form 3

Distance-time Graphs

Area under velocity-time graph
Consider a body with uniform or constant acceleration for time ‘t’ seconds;

Distance travelled= average velocity × t
= (0 + v/2) × t
= ½vt
This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under 't' seconds.
Example 8
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;
i. Calculate the total distance moved by the car.
ii. Find the acceleration of the car at each stage.

Solution

1. a. From the graph, total distance covered= area of (A + B + C)
   = (½ × 10 × 20) + (½ × 6 × 20) + (5 × 20)
   = 100 + 60 + 100
   = 260m
   Also the area of the trapezium gives the same result.
2. 
   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 50Hz i.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 10×0.02= 0.2 seconds.

Example 9
a. 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

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Example 10
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 × 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;

1. v = u + at
2. s = ut + ½ at²
3. v²= u² + 2as

Example 11
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) + 2 × 10 × 320
= 3600 + 6400
= 10,000
Therefore v = (10,000)^1/2
v= 100m/s

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Example 12
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.

Example 13
A body is uniformly accelerated from rest to a final velocity of 100m/s in 10 seconds. Calculate the distance covered.
Solution

s = ut + ½ at²
= 0 × 10 + ½ × 10 × 10²
= 1000/2 = 500m

Motion Under Gravity

1. Free Fall

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

1. v = u + g t
2. s = ut + ½ gt²
3. v²= u + 2gs

2. Vertical projection

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

1. v = u - gt
2. s = ut -½ gt²
3. v² = u - 2gs
   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 time taken to fall back to its point of projection. Using eq. 2 then, displacement =0

1. 0 = ut - ½gt²
2. 0 = 2ut - gt²
3. t (2u-gt) = 0
   Hence, t = 0 or t = 2u/g
   t = o corresponds to the start of projection
   t = 2u/g corresponds 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

1. v2= -2gs;- 0 = u² - 2gHmax, therefore
2. 2gHmax = u²
3. 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
Therefore v = u or v = ± u

Example 14
A stone is projected vertically upwards with a velocity of 30m/s from the ground. Calculate,
a. The time it takes to attain maximum height
b. The time of flight
c. The maximum height reached
d. The velocity with which it lands on the ground. (take g=10m/s)

Solution

1. Time taken to attain maximum height
   T = u/g = 30/10 = 3 seconds
   
2. The time of flight
   T = 2t = 2 × 3 = 6 seconds
   Or T = 2u/g = 2 × 30/10 = 6 seconds.
   
3. Maximum height reached
   Hmax = u²/2g = 30 × 30/2 × 10 = 45m
   
4. Velocity of landing (return)
   v²= u² - 2gs, but s = 0,
   Hence v² = u²
   Therefore v = (30 × 30)^1/2 = 30m/s

3. 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 'R' at a time 't' is given by s = ut + ½at²
Taking u = u and a = 0 hence R = ut, is the horizontal displacement and h = ½gt² is the vertical displacement.
NOTE: The time of flight is the same as the time of free fall.

Example 15
A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s. Calculate,
a. The time taken by the ball to strike the ground.
b. The distance from the foot of the cliff to where the ball strikes the ground.
c. The vertical velocity at the time it strikes the ground. (take g=10m/s)
Solution

1. h = ½ gt²
   20= ½ × 10 × t²
   40 = 10t²
   t² = 40/10 = 4
   t = 2 seconds
   
2. R = ut
   = 10 × 2
   = 20m
   
   
3. v = u + at = gt
   = 2 × 10 = 20m/s