Physics Form 3 Notes : CHAPTER ONE – LINEAR MOTION

PHYSICS FORM THREE

CHAPTER ONE

LINEAR MOTION

Introduction

Study of motion is divided into two:

1. Kinematics
2. Dynamics

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

1. Displacement

Distance moved by a body in a specified direction is called displacement. It is denoted by the letter ‘s’ and has both magnitude and direction. Distance is the movement from one point to another. The SI unit for displacement is the metre (m).

1. 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

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

   Solution

   Total distance covered = 10 + 90 = 100 m

   Total time taken = 4 + 10 + 60 = 74 seconds

   Therefore average speed = 100 / 74 ≈ 1.35 m/s

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

   Solution

   Speed = 180 × 1000 / 3600 = 50 m/s

   Distance covered = speed × time

   = 50 × 30

   = 1,500 m

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

   Solution

   Speed: 360 km/h = 360 × 1000 / 3600 = 100 m/s

   Time = distance / speed

   = (3000 × 1000) / 100

   = 30,000 seconds.

1. 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

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

      Solution

      1. Average speed: total distance travelled / total time taken

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

      = 1200 / 180

      = 6.67 m/s

      1. Average velocity: total displacement / total time

         = (800 – 400) / 180

         = 400 / 180

         = 2.22 m/s due North

      2. Change in velocity = final velocity – initial velocity

         = (800 / 100) – (400 / 80)

         = 8 – 5

         = 3 m/s due North

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

      Solution

      Initial velocity (u) = -10 m/s

      Final velocity (v) = 10 m/s

      Therefore change in velocity = v – u

      = 10 – (-10)

      = 20 m/s

2. Acceleration

   This is the change of velocity per unit time. It is a vector quantity symbolized by ‘a’.

   Acceleration ‘a’ = 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 = 20 m/s

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

      Therefore a = (v – u) / t

      = (40 – 20) / 10

      = 2 m/s²

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

      Solution

      Initial velocity = 180 km/h = 50 m/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

b)

c)

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

= ½ v t

This is equivalent to the area under the graph. The area under a velocity-time graph gives the distance covered by the body under ‘t’ seconds.

Example

A car starts from rest and attains a velocity of 72 km/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;

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

   Solution

   

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

      = (½ × 10 × 20) + (5 × 20) + (½ × 6 × 20)

      = 100 + 100 + 60

      = 260 m

      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 50 Hz 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.

Examples

1. 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 = 5 cm

   Frequency of the ticker-timer = 50 Hz

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

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

2. The tape below was produced by a ticker-timer with a frequency of 100 Hz. 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:

v = u + at

s = ut + ½ at²

v² = u² + 2as

Examples

1. 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 + 2 × 10 × 320

   = 3600 + 6400

   = 10,000

   Therefore v = √10,000

   v = 100 m/s

2. A body whose initial velocity is 30 m/s moves with a constant retardation of 3 m/s². Calculate the time taken for the body to come to rest.

   Solution

   v = u + at

   0 = 30 – 3t

   3t = 30

   t = 10 seconds

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

   Solution

   s = ut + ½ at²

   = 0 × 10 + ½ × 10 × 10²

   = 0 + ½ × 10 × 100

   = 500 m

Motion under gravity

1. Free fall

   The equations used for constant acceleration become:

   v = u + g t

   s = ut + ½ gt²

   v² = u² + 2gs

2. Vertical projection

   Since the body goes against the force of gravity, the following equations hold:

   v = u – g t ……………1

   s = ut – ½ gt² ……2

   v² = u² – 2gs …………3

   Note: Time taken to reach maximum height is given by:

   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, displacement = 0:

0 = ut – ½ gt²

0 = 2ut – gt²

t (2u – gt) = 0

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

t = 0 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, H_max, is attained when v = 0 (final velocity). Hence:

v² = u² – 2gH_max; 0 = u² – 2gH_max, therefore

2gH_max = u²

H_max = u² / 2g

Velocity to return to point of projection

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

v² = u² – 2gs hence v² = u²

Therefore v = u or v = ±u

Example

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

1. The time it takes to attain maximum height
2. The time of flight
3. The maximum height reached
4. The velocity with which it lands on the ground. (take g = 10 m/s²)

   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

      H_max = u² / 2g = 30 × 30 / 2 × 10 = 45 m

   4. Velocity of landing (return)

      v² = u² – 2gs, but s = 0,

      Hence v² = u²

      Therefore v = √(30 × 30) = 30 m/s

5. 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 + ½ at².

   Taking u = u and a = 0, hence R = ut is the horizontal displacement and h = ½ gt² 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 20 m high with a horizontal velocity of 10 m/s. Calculate:

   1. The time taken by the ball to strike the ground
   2. The distance from the foot of the cliff to where the ball strikes the ground
   3. The vertical velocity at the time it strikes the ground. (take g = 10 m/s²)

      Solution

      1. h = ½ gt²

         20 = ½ × 10 × t²

         20 = 5 t²

         t² = 20 / 5 = 4

         t = 2 seconds

      2. R = ut

         = 10 × 2

         = 20 m

      3. v = u + at = g t

         = 10 × 2 = 20 m/s