CHAPTER ELEVEN: MOTION – PHYSICS S1-S2 NOTES

11.1  Speed velocity and Acceleration

(a)  Speed

Speed is defined as the rate of change of distance moved with time.

OR  Speed is the distance traveled in a unit time.

Mathematically, it is expressed as:  Speed =

S.I unite of speed is m/s or ms^-1.

(b)  Displacement  

Displacement is defined as distance moved in a specified direction.

For example if a body moves a long a straight line in a given direction such as 100m due east or 50m due north.

(c)  Velocity  

 Velocity is defined as the rate of change of displacement with time.

Or  Velocity is the rate of change of distance moved with time in a specified direction.

Or   Velocity is defined as speed in a specified direction.

The S.I unite of velocity is ms^-1 or m/s.

 Difference between speed and velocity

Velocity is a vector quantity whereas speed is a scalar quantity.

NB:  Vector quantity is a quantity that has both magnitude and direction.

Scalar quantity is a quantity that magnitude only.

Uniform Velocity  

Uniform velocity is the velocity when the rate of change of displacement is constant.

Non-uniform velocity

Non-uniform velocity
is the velocity when the rate of change of displacement is not constant.

Acceleration

Definition  Acceleration is defined as the rate of change of velocity with time.

Acceleration  =

=  

The S.I unit of acceleration

The SI unit of acceleration  = m/s² or ms^-2 . Read as metres per sec squared.

When the velocity of a body is changing, the body is said to be accelerating.

Acceleration is:  Positive  – if the velocity is increasing and

Negative  – if the velocity is decreasing.

 A negative acceleration is called deceleration or retardation.

Uniform Acceleration  

Uniform acceleration is the acceleration when the rate of change of velocity is constant.

11.2  The Equations of Linear Motion

The following are the equations for uniformly accelerated body.

Notes: 1.  When using the equations, it is necessary to bear in mind that u, v, a, and s are vectors. If, say, the positive direction is taken to be up, then:

1. The velocity of a body which is moving down
   (i.e in opposite direction) is negative.
2. Points below the starting point have negative values of s.
3. Downward directed accelerations are negative.

2.  This facts will help a learner to understand a velocity-time graph with negative values of velocity.

Derivation of the Equations of Motion

Suppose that a body is moving with constant acceleration a and that in a time interval t its velocity increases from u to v and its displacement increases from 0 to s. Then, from the definition of acceleration we have:

Acceleration  = Rate of change of velocity

Acceleration  =

=

a  =

v  = u + at ……………………….  1

Since the acceleration is uniform,

Average velocity  =

= and therefore

Displacement = Average velocity x Total time

s  =

s  = ½(v + u)t …………….………….  2

Eliminating v from equation gives equation 3. This is done by substituting equation 1

(v = u + at) in equation 2.

  s  =

=

=

  s  = ut + at^{2 ………………………………..}  3

Eliminating t from equation (2) gives equation (4).

This is done by substituting  t = from equation (1) in equation (2).

I.e. s  = x

  s  =

  2as  = v² – u²

  v²  = u² + 2as ………………………….  4

Worked Examples

1. An object starts from rest and moves with an acceleration of 10 m/s² for four seconds.

Calculate:  (a)  The velocity

(b)  The distance traveled

 Solution:  (a)  u = 0, a = 10 ms^-2, t = 4 s, v = ?

Using the formula   v  = u + at

= 0 + 10 x 4

  v  = 40m/s

(b)  Using the formula   s  = ut + ½at²

= 0 x 4 + ½ x 10 x 4 x 4

= 0 + 80 m

\
s  = 80 m

1. A car moving with uniform acceleration of 4 m/s² increases its velocity from 20 m/s to 60 m/s, Calculate: (a )  the total time taken during this change

   ( b)  the total distance moved

Solution: a = 4 m/s², u = 20 m/s, v = 60 m/s, t = ?, s = ?

(a)  Using the formula v  = u + at

60  = 20 + 4 x t

60  = 20 + 4t

4t²  =
60 – 20

4t²  = 40

 =

\
t  = 10 s

(b)  Using the formula   s  = ut + ½at² = 20 x 10 + ½ x 4 x 10 x 10

= 200 + 200

\
s  = 400 m

3.  A body traveling with a velocity of 10m/s is uniformly retarded to rest in 5 seconds

 Find  (a)  its acceleration.

 (b)  the distance traveled during the retardation.

Solution:   u = 10 m/s, v = 0, t = 5 s, a =?, s = ?

 (a)  Using the formula v  = u + at

0  = 10 + 5

 =

\a  = -2 m/s²

The minus sign means that the body is accelerating in the opposite direction to its initial velocity.

(b)  Using the formula:   v²  = u² + 2as

  0  = 10² + 2 x 2 x s

0  = 100 + 4s

4s  = 100

 =

\
s  = 25 m

 11.21  Changing velocity from km/h to m/s

Velocity given in km/h can be changed to m/s. This is simply done by applying the formula for speed.

I.e  Speed  =

Steps followed

• Interpret the value of the speed given.

E.g. x km/h   means that the body travels x km in 1 hour.

• Change the numerical value of the velocity or speed to metres and also change 1 hour to seconds.
• Divide the distance in metres by the time in seconds.

Example  

Change the speed 36 km/h to m/s.

The speed 36 km/h,  means that the body travels 36 km in 1 hour.

To change the speed from km/h to m/s, we change the 36 km to m and 1 hour to seconds.

Changing 36 km to m: 1 km  = 1000 m

36 km  = x

  x  = = 36 000 m

Changing 1 hour to seconds:  1 h  = 60 min

= (60 x 60) s

= 3600 s

 Now we substitute the distance in m and time in seconds in the formula:

 Speed  =
= = 10 m/s

Self-Check  11.1

1. An object starts from rest and moves with an acceleration of 10 m/s² for four seconds.

Calculate:  (a)  The velocity

   (b)  The distance traveled .

1. A car moving with uniform acceleration of 4 m/s² increases its velocity from 20 m/s to 60 m/s, Calculate:  (a)  the total time taken during this change

   (b)  the total distance moved

3.  A body traveling with a velocity of 10m/s is uniformly retarded to rest in 5 seconds

 Find:  (a)  its acceleration.

 (b)  the distance traveled during the retardation.

4.  Change the following speeds from km to m/s.

(a) 72 km/h (b)  108 km/h  (c)  180 km/h  (d)  288 km/h

11.3  Graphical Representation of Linear Motion

Graphs can be used to represent the motion of a body which is moving in a straight line.

(a)  Types of Graphs of Linear Motion

There are two types of graphs namely:

• Distance-time graph and
• Velocity-time graph.

(b)  Distance-Time graph

A distance-time graph is the type of graph obtained by plotting distance/displacement travelled against time taken.

The common graphs are for a body:

• Moving with a uniform velocity
• Non-uniform velocity and
• At rest.

By definition, velocity is rate of change of displacement and therefore the slope of a graph of displacement against time represents velocity.

(i)  Uniform velocity  (ii)  Non-uniform velocity

Velocity  = Velocity  =  

= Velocity at A  =

I.e Velocity  = Gradient of the line OA = Gradient at A

(iii)  A body at rest  (iv)  A body on return journey

However, these graphs can be combined to form a single graph with the different stages of the motion as shown below.

OA  – The slope is positive

• The body is moving constant velocity during this period
  

AB  – The slope is zero i.e the velocity is zero

• The body is stationary

BC  – The slope is negative

• The body is moving constant velocity in the opposite direction during this period
  

Velocity-Time graph  

A velocity-time graph is the type of graph obtained by plotting velocity against time.

The common graphs are for a body:

• moving with uniform velocity (constant velocity)
• Non-uniform velocity,
• Uniform acceleration,
• Non-uniform acceleration and
• Uniform deceleration

By definition, acceleration is rate of change of velocity and therefore, the slope of a graph of velocity against time represents acceleration.

(i)  Uniform velocity, (ii)  Uniform acceleration

(iii)  Uniform deceleration (iv)  Non-uniform acceleration

Acceleration  =    Acceleration =  

=  Acceleration at A  =

Acceleration  = Gradient of the line OA  = Gradient at A

However, the graphs for uniform acceleration, velocity and deceleration can be combined to represent a linear motion for a body starting from rest or at a certain initial velocity. Then the total distance travelled is equal to the area under the graph and can be calculated using equations of motion or using appropriate formulae.

A velocity-time graph for a body that travelled a given distance and returned and stopped on the way.

 Notes: 1.   When interpreting a graph, it is necessary to bear in mind that u, v, a, and s are

  vectors.

 If, say, one direction is taken to be positive, then:

(i)  The velocity of a body which is moving in the opposite direction is negative.

(ii)  Points below the starting point have negative values of s.

(iii)  Downward directed accelerations are negative.

(iv)  The magnitude and the direction of the slope of the graph give the magnitude and direction of the acceleration.

2.  These facts will you to understand the interpretation of the above velocity-time graph where there are negative values of velocity.

Worked Examples

1.  A car starts from rest and accelerates uniformly for time t₁ seconds to attain a maximum velocity v ms^-1. It maintained this velocity for t₂ seconds, then the brakes were applied and the car uniformly retarded to a rest for t₃ seconds.

(i)  Represent the above motion in a velocity-time graph.

(ii)  On your graph show by shading the total distance travelled.

Solution:  Initial velocity = 0 (from rest), Final velocity = v ms^-1

The velocity-time graph for the motion

  0 t₁ (t₁+t₂) (t₁+t₂+t₃)

Total distance travelled  = Area under the graph

= Area of the trapezium

= ½(a + b)h

Or  Total distance travelled  = Area A + Area B + Area C

= ½bh + lw + ½bh

2.  A car starts from rest and is accelerated uniformly at the rate of 5 ms^-2 for time 10 seconds to attain a maximum velocity. It then maintained this velocity for 40 seconds. The brakes were then applied and the car uniformly retarded to rest in 5 seconds.

 Calculate:  (i)  the maximum velocity attained.

(ii)  Represent the motion on velocity-time graph.

(iii)  Find the total distance travelled in metres.

Solution  (i)  u = 0, v = ?, a = 5 ms^-2, t = 10 s

v  = u + at

 = 0 + 5 x 10

\ v  = 50 ms^-1

  (ii)  The velocity-time graph for the motion

1. Method I  a = 50 – 10 = 40, b = 55, h = 50  

   Total area  = Area trapezium OABC

 = ½(a + b)h

 = ½(AB + OC) x AD

 = ½(50 – 10 + 55) x 50

 = ½(40 + 55) x 50

 = ½ x 95 x 50

 = 2375 m

\ Total distance travelled  = 2375 m

Method II  Total distance travelled  = Area A + Area B + Area C

= ½bh + lw + ½bh

= ½ x 10 x 50 + (50 – 10) x 50 + ½ x (55 – 50) x 50

= 250 + 40 x 50 + ½ x 5 x 50

= 250 + 2000 + 125

= 2375 m

\Total distance travelled  = 2375 m

3.  A car travels with a uniform velocity of 30 ms^-1 for 5 seconds. The brakes are then applied and the car comes to a rest with a uniform retardation in a further 10 seconds.

(i)  Draw a sketch graph of velocity-time graph.

(ii)  Show by shading the total distance travelled by the car.

(iii)  Find the distance travelled by the car after the brakes are applied.

 Solution:  u = 30 ms^-1, t₁ = 5 s, t₂ = 10 s

(i) Velocity-time graph for the motion

(ii)  The distance travelled after the

brakes were applied

 = Area B

 = ½bh

 = ½(15 – 5) x 20

 = ½ x10 x 30

 = 150 m

4.  A body starts from rest and accelerates at 5 ms^-2, for 5 seconds. Its velocity remained constant at the maximum value so reached for 12 seconds and it finally came to a rest with uniform retardation after another 4 seconds.

 Find by graphical method:

(i)  The distance travelled during each stage of the motion.

(ii)  The average velocity for the whole journey.

 Solution  u = 0,   v = ?, a = 5 ms^-2  t₁ = 5 s, t₂ = 12 s, t₃ = 4 s

v  = u + at = 0 + 5 x 5 = 25 ms^-1

Velocity-time graph for the motion

Dist. travelled at the first stage = Area A

  = ½bh

  = ½ x 5 x 25

  = 62.5 m

\ Distance travelled   = 62.5 m

Distance travelled at the second stage  = Area B

= lw

= (17 – 5) x 25

= 12 x 25

= 300 m

Distance travelled at the third stage = Area C

= ½bh

= ½ x (21 – 17) x 25

= ½ x 4 x 25

\ s  = 50 m

(ii)  Average velocity =

 =

 =

 =

\ Average velocity = 19.64 ms^-1

Self-Check  11.2

1.  A body moves with uniform acceleration if

A. its momentum remains constant

B. it covers equal distances in equal times

C. the velocity changes by equal amount in equal times  

D. the net force on the body is zero

2.  A cyclist traveling at a constant acceleration of 2ms^-2 passes through two points A and B in a straight line. If the speed at A is 10ms^-1 and the points are 75m apart, find the speed at B.

A. 15.8ms^-1 B. 17.3ms^-1 C. 20.0ms^-1 D. 400.0ms^-1

3.  A body moves with a uniform acceleration of Pms^-2. If its initial velocity is xms^-1 and it travels for t s to attain a final velocity of yms^-1, find the value of p in terms of x, y and t.

A. B. C. D.

4.  A body is said to be moving with a constant velocity if;

(i)  Its momentum remains constant

(ii)  It covers equal distances in equal times

(iii)  The velocity changes by equal amount in equal times

(iv)  The net force on the zero

 A. All B. (i), (ii) and (iv) C. (i) and (ii) only  D. (ii) only

5.  A car of mass of mass 1200 kg moving with a constant velocity of 60 ms^-1 is retarded uniformly to rest in 12 sec .Calculate the retarding force.

A. (1200 x 12) N  B (1200 x 5) N  C. (1200 x 10) N  D. (1200 x 60) N

6.  The gradient of a velocity-time graph represents the

  A. Speed of the body B. Velocity of the body

C. Acceleration of the body D. The distance covered by the body

7.  A body is said to be moving with uniform velocity when the rate of change of

A. acceleration with time is constant B. velocity with time is constant

C. distance with time is constant D. displacement with time is constant.

8.  Which one of the above sketches represents uniformly accelerated motion?

9.  A lift accelerates uniformly from rest for 3s. It then moves at uniform velocity for 15s then

decelerates uniformly for 2s before coming t rest. Which of the following velocity-time graph represents the motion of the lift.

10.  Figure 4 shows a velocity-time graph for a moving body. Which one of the following statements is true about the motion of the body?

A. Velocity of the body is constant between O and A.

B. Velocity of the body is constant between A and B.

C. The body is accelerating between A and B.

D. The body is not accelerating between O and A.

11.  The graph in figure 1 describes the motion of particle. Between which points is the particle at rest?

A. O and P.  

B. P and Q.

C. Q and R.

D. R and S.

12.  A car starts from rest and accelerates uniformly at the rate of 2ms^-2 from 1/4 of a minute. Find the velocity of the car after this time.

A. 0.5ms^-1 B. 12ms^-1 C. 15ms^-1  D. 30ms^-1

13.  The in the figure shows a speed-time graph for a body.

Calculate the distance traveled during retardation

A. 20m B. 40m C. 80m D. 100m

14.  A boy throws a ball in the air and it goes up and falls back to his hand. Which one of the following sketches of velocity-time graph represents the motion of the ball up to the time it is received back?

15.  The velocity-time graph for a car is as shown in fig. Find the total distance the car travels.

A. 2.08×10³m  

B. 3.0×10³m  

C. 4.0×10³m  

D. 7.0×10³m

16.  Which of the following best describes the motion represented by the velocity-time graph shown in the diagram?

A. Decelerated

B. Uniformly accelerated motion

C. Non-uniformly accelerated motion

D. Uniform velocity motion

17.  A car is uniformly accelerated from rest and after 10s acquires a speed of 20ms^-1. How far does it move during the eleventh second?

A. 20m  B. 21m C. 100m    D. 121m

18.  Which one of the following is not true about a body moving with a constant velocity?

A. its acceleration is zero B. Its momentum is constant

C. Its kinetic energy is constant D. There is a resultant force on it

19.  A car travelling at 20ms^-1 is brought to rest in 10s. Find the distance it travels

A. 100m B. 200m C. 300m D. 400m

20.  Use
the velocity-time graph in the figure to find the distance over which there is deceleration.

A. 4m

B. 6m

C. 8m

D. 10m

SECTION B

21.  (a)  What is meant by the following:

(i)  uniform velocity  (ii)  uniformly accelerated motion  ?

(b)  A body starts from rest and reaches a speed of 5m/s after travelling with uniform acceleration in a straight line for 2 seconds. Calculate the acceleration of the body.

22.  A body starts from rest and moves with uniform acceleration of 2m/s in a straight line.

(a)  What is its velocity after 5 seconds?

(b)  How far has it travelled in this time?

(c)  How long will he body be after the starting point?

23.  (b)  The table below shows the variation of velocity with time for a body, which has

been thrown vertically up wards from the surface of a planet.

(i)  What does the negative velocity mean?

(ii)  Plot a graph of velocity against time.

(iii)  Use the graph in b (ii) to find the acceleration due to gravity on the planet.

(iv)  Use the graph in b (ii) to find the total distance traveled.

24.  (a)  What is meant by acceleration?

(b)

The above figure shows a speed-time graph of a cyclist.

(i)  Find the acceleration of the cyclist between A and B.

(ii)  Describe the motion of the cyclist between B and C

(iii)  Explain what is happening along CD.

(iv)  Calculate the distance travelled by the cyclist during the first 10 seconds

26.  (a)  What is the difference between speed and velocity?

(b)  The graph in the figure shows the variation of distance with time for a body. Describe the motion of the body.

11.4  Ticker-Tape Timer (ticker-tape vibrator)

This is a device used to investigate speed, velocity and acceleration of a moving body. It consists of a steel strip which vibrates at mains frequency, f, making dots on paper tape which is pulled by the body. The time interval between any two successive dots is called a tick and is equal to the reciprocal of the frequency in seconds.

Calculations

To calculate speed, velocity or acceleration, the following steps are followed.

First determine: (i) The distance between the reference points (dots) on the ticker tape

1. The tick using the formula

One tick = second.

(iii)  The time taken to make the dots from the formula

Total time taken  = One tick x

(a)  To calculate velocity

1.  The diagram below shows equally spaced dots on a ticker tape.

If the vibrator of the ticker time has a frequency of 50 Hz, find the average speed of the trolley in m/s.

Solution:  Frequency  = 50Hz, One Tick = ?

One tick  =  =  = 0.02 s

Total distance covered = 50 cm = m = 0.5 m

Total time taken  = One tick x

= 0.02 x 6

t  = 0.12 s

Using Speed = = = 4.166  = 4.2 ms^-1

(b)  To calculate the acceleration

1.  The tape shown in the diagram below was made by a trolley moving with a constant acceleration. If the frequency of the ticker-timer is 100 Hz, find the acceleration in ms^-2.

Solution: F = 100 Hz  One tick = = = 0.01 s

 Initial Velocity, u, at X:   Distance, s  = 2 cm =  = 0.02 m

 Time taken  = One tick x No. of spaces between the dots

 = 0.01 x 2

 = 0.02 s

Initial velocity, u  =
=  = 1 ms^-1

 Final Velocity, v, at Y:   Distance, s  = 4 cm =  = 0.04 m

 Time taken  = One tick x No. of spaces between the dots

 = 0.01 x 2

 = 0.02 s

Initial velocity, u  =
=  = 2 ms^-1

Time taken from X to Y is calculated from the formula

Time taken  = One tick x No. of spaces between XY.

 = 0.01 x 5

t  = 0.05 s

The acceleration is calculated from the formula:

Acceleration  =  

a  = = = = 40 ms^-2

2.  Calculate the acceleration, in ms^-2, for the motion shown in the diagram below.

(Take frequency = 50Hz)

Solution: F = 50 Hz  One tick = =  = 0.02 s

 Initial Velocity, u, at P:   Distance, s  = 2 cm =  = 0.02 m

 Time taken  = One tick x No. of spaces between the dots

 = 0.02 x 2

t  = 0.04 s

Initial velocity, u  =
=
= 0.5 ms^-1

 Final Velocity, v, at Q:   Distance, s  = 5 cm  =  = 0.05 m

 Time taken, t  = One tick x No. of spaces between the dots

 = 0.02 x 2

 = 0.04 s

Time taken from P to Q is calculated from the formula

 Time taken, t  = One tick x No. of spaces between P and Q

 = 0.02 s x 4

t  = 0.08 s

Initial velocity, v  =
=  = 1.25 ms^-1

The acceleration is calculated from the formula

Acceleration  =  

a  = =  =
= 9.375 ms^-2

(c) To Describe Motion

The figure below shows dots made on a ticker tape pulled by a trolley through a ticker timer.

Describe the motion of the trolley if the frequency is 50Hz?

To describe motion from a sample of ticker tape, you find the speeds at the two stages of the motion and their respective time taken. Then relate the speeds to the time taken for the two stages as shown below.

Speed at the first stage

Frequency, f = 50 Hz,  Tick = = = 0.02 s, s = 4 cm,

Time taken  = One tick x No. of spaces

= 0.02 x 2

= 0.04 s

 Speed  = = =100 cms^-1

Speed at the second stage

s = 8 cm,  time taken  = One tick x No. of spaces

= 0.02 x 5

= 0.1 s

 Speed  =
= = 80 cms^-1

Description of the motion

In the first 0.04 s, the body was traveling with a speed of 100 cms^-1. The body then decelerated and the speed reduced to 80 cms^-1 in the next 0.1 s.

Self-Check  11.3

1.  A ticker timer is connected to the mains supply of frequency 40Hz. Find the time it takes to print three consecutive dots.

A. 0.08 s B. 0.25 s C. 0.050 s  D. 0.75 s

2.  The ticker tape shown in the figure was pulled through a ticker timer, which makes 50 dots per second.

The speed at which the tape was pulled is

A. 10cm s^-1  B. 25cm s^-1 C. 50cm s^-1 D. 100cm s^-1

3.  A tape is pulled through a ticker-timer, which has a frequency of 50Hz. If the distance

between successive dots is 2cm, calculate the speed of the body

A. 0.01cms^-1 B. 50cms^-1 C. 100cms^-1 D. 250cms^-1

4.  The equally spaced dots on the ticker tape above are made by a vibrator of

frequency 50 Hz. The speed of the tape in m/s is

A. 0.1 B. 0.01 C. 0.001 D. 1

5.  The tape shown in the diagram below was made by a trolley moving with a constant

acceleration. If the frequency of the ticker-timer is 50 Hz, find the acceleration in m/s².

A. 75 ms^-2    B. 2 ms^-2    C. 5 ms^-2  D. 12.5 ms^-2  

6.  Calculate the acceleration, in ms^-2, for the motion shown in the diagram below.

(Take frequency = 100Hz)

A. 75 ms^-2    B. 2 ms^-2    C. 5 ms^-2  D. 0.04 ms^-2  

11.5  Motion under Gravity

Types of Motion

There are different forms of motion under gravity, depending how the motion is started.

(i)  Free-falling objects  – If an object is released from rest (i.e. when the

initial velocity is zero) and allowed to fall.

(ii)  Projectile motion  – If an object is thrown (projected) at an angle with an

initial velocity greater than zero.

1. Free-falling objects
   

Objects released from rest at a height above the surface of the earth fall downwards. This is because of force of gravity pulling them downwards.

An object thrown vertically upwards with an initial velocity greater than zero moves with a negative acceleration since the motion is away from the earth surface. The velocity reduces until it becomes zero at the maximum height reached. At this point the object is momentarily at rest and then begins to accelerate downwards with a positive acceleration. The velocity increases until it hits the ground. This acceleration arises from the force of gravity acting on the body, and is called acceleration due to gravity (symbol g).

If air resistance or friction is neglected, all objects near to the surface of the earth, (regardless of their mass) accelerate at the same rate called acceleration due to gravity. The acceleration due to gravity (symbol g) is about
9.81 ms^-2. But the value varies slightly from place to place. See chapter five. For simple calculations g is usually rounded up to 10 ms^-2. This means that objects falling freely under gravity increase their velocity by about ms^-1 every second.

Calculations for free-falling objects

The equations of motion, apply to objects moving vertically downwards or upwards under the force of gravity, except that the acceleration a is replaced by the acceleration due to gravity, g.

Distance moved by a free-falling body

Distance moved by a body falling freely from rest is related to time of fall by the

formula:  s = ½g t²

The
formula is derived from the third equation of motion

s = ut + ½at²

Where u = 0  (i.e. from rest)

  a = g  acceleration due to gravity

NB:  An object moving vertically upwards has a negative acceleration or retardation. Therefore the acceleration due to gravity, g = ^–10 ms^-2 while for an body moving downwards, the acceleration due to gravity is positive, g = 10 ms^-2

Worked Examples

1.  A coconut fruit falls from the coconut tree and takes 5 seconds to hit the ground.

Calculate:  (a)  its velocity.

(b)  the total distance traveled. (Take g = 10m/s²)

  Solution  (a)   g = 10m/s², u = 0 ms^-1, t = 5 s

    v  = u + at

= 0 + 10 x 5

 \ v  = 50 ms^-1

(b)  Using the formula s = ut +½at²

 = (0 x 5) +½ x 10 x 5²

 = 0 + 5 x 25

\ s  = 125 m

 2.  A stone is dropped from a top of a cliff. If it takes 8 seconds to hit the ground,

calculate:  (a)  the height of the cliff.

 (b)  the velocity at impact. (Take g = 10 ms^-2)

   Solution  (a)  Using the formula g = 10m/s², u = 0 ms^-1, t = 4 s  

s  = ut +½gt²

 = (0 x 4) + x 10 x 4²

 = 0 + 5 x 16

\ s  = 80 m

(b)     v  = u + at

= 0 + 10 x 4

\ v  = 40 ms^-1

 3.  A stone is thrown vertically upwards with an initial velocity of 40 ms^-1.

Calculate:  (a)  the maximum height reached.

 (b)  the time taken to reach the maximum height.

(Take g = 10 ms^-2 and neglect air resistance)

Solution

 u = 40 ms^-1, v = 0 ms^-1(At maximum height the stone is momentarily at rest),

a = g = -10 ms^-2, s = ?

(a)   v²  = u² + 2as

v²  = u² + 2gs

0²  = 40² + 2 x -10 x s

0  = 1600 – 20s

20s  = 1600

s  = 80 m

(b)  u = 40 ms^-1, v = 0 ms^-1, a = g = -10 ms^-2, s = ?

v  = u + at

v  = u + gt

0  = 40 + -10 x t

0  = 40 – 10t

10t  = 40

\ t  = 4 s

4.  A ball is thrown vertically upwards with an initial velocity of 20 ms^-1. Neglecting air resistance, calculate:

 (a)  the maximum height reached,

(b)  the time taken before it reaches the ground. (Take g = 10 ms^-2)

 Solution  u = 20 ms^-1, v = 0 ms^-1(At maximum height the stone is momentarily at rest),

a = g = -10 ms^-2, s = ?

(a)   V²  = u² + 2as

V²  = u² + 2gs

0²  = 20² + 2 x -10 x s

0  = 400 – 20s

20s  = 400

\ s  = 20 m

(b)  u = 20 ms^-1, v = 0 ms^-1, a = g = -10 ms^-2, t₁ = ?

v  = u + at

v  = u + gt

0  = 20 + -10 x t₁

0  = 20 – 10t₁

10t₁  = 20

t₁  = 2 s

Time taken in the return journey, t₂ = ? u = 0 ms^-1, v = 0 ms^-1, a = g = 10 ms^-2, s = 20 m

Using the formula   s   = ut +½at²

20  = (0 x t₂) +½ x 10 x t₂²

20  = 5t₂²

Ö4  = Öt₂²

t₂  = 2 s

Time taken before it reaches the ground  = t₁ + t₂ = 2 + 2 = 4 s

NB:  Note that the time taken to reach the maximum height is equal to the time taken to return to the ground.

Further Examples

1.  A stone is thrown vertically upwards with an initial velocity of 30 ms^-1 from a tower 20 m high. Neglecting air resistance, calculate:

 (a)  The time taken to reach the maximum height.

(b)  The maximum height reached.

(c)  The total time taken which elapses before it just hits the ground

(Take g = 10 ms^-2)

Solution

 (a) u = 30 ms^-1, v = 0 ms^-1
(At the maximum height the stone is momentarily at rest),

a = g = -10 ms^-2, t = ? s = ?

v  = u + gt

0  = 30 + -10 x t

0  = 30 – 10t

10t  = 30

t₁  = 3 s

(b)   v²  = u² + 2as

v ² = u² + 2gs

0²  = 30² + 2 x -10 x s

0  = 900 – 20s

20s  = 900

\ s  = 45 m

(c)  Total height = Height of cliff + Maximum height reached

= 20 + 45

  \  Total height = 65 m

Time to fall from the maximum height to the ground =?, g = 10 ms^-2

Using the formula:   s   = ut + ½at²

  65  = (0 x t) +½ x 10 x t²

  65  = 5t₂²

  Ö13  = Öt²

t  = 3.60 s

Total time taken  = +

= 3 + 3.60

\ Total time taken  = 6.60 s

2.  A small marble chip of mass 5 g is dropped from the top of a cliff and takes 2.5 s to reach the sandy beach below it. Calculate:

 (a)  (i)  the velocity with which it strikes the sand.

(ii)  the kinetic energy with which it hits the sand.

(b)  The height of the cliff.

(c)  If the marble chip penetrates the sand to a depth of 12.5 cm, calculate:

(i)  its average retardation.

(ii)  the retarding force. (Take g = 10 ms^-2)

Solution:  (a)  u = 0 ms^-1, v = ?, a = g = 10 ms^-2, t = 2.5 s,

 (i)   v  = u + at

v  = u + gt

= 0 + 10 x 2.5

= 0 + 25

= 25 ms^-1

1. v = 25 ms^-1, m = 5 g = kg, K.E = ?

K.E  = ½mv²

= x x 25²

=

=

K.E = 1.56 J

(b)  g = 10m/s², u = 0 ms^-1, t = 2.5 s, s = ?, v = 25
ms^-1

s  = ut +½gt²

= (0 x 2.5) +½ x 10 x 2.5²

= 0 + 5 x 2.5 x 2.5

= 31.25 m

(c)  (i)  Depth of penetration, s = 12.5 cm m, a = ?

u  = initial velocity with which the marble starts to penetrate the sand

= velocity before it just hits the sand

= 25 ms^-1

v  = 0 (after penetration the marble was brought to rest)

Using the formula   v²  = u² + 2as

v²  = u² + 2gs

  a
 =

=

=
= 2500 ms^-2

m = kg, a = 2500 ms^-2, F = ?

F = ma

= x 2500

=

= 5 N

Experiment 5.31  

To Determine Acceleration due to Gravity, g, in a place

Apparatus/Requirements

2 G-clamps, a retort stand, 2 m ticker tape, a ticker timer, a standard mass of 100 g, a soft material.

Procedure

• Clamp a ticker timer about 2 m or more at the edge of a table so that there is a clear drop below it as shown in figure 5.1 below.

Figure 5.1

• Pass a 2 m ticker tape through the paper guides and under the carbon disc.
• Place a soft material on the floor for a 100 g mass to fall on.
• Attach one end of the tape to 100 g.
• Switch on the ticker timer and allow the 100 g mass to drop to the ground.
• Construct a tap chart using two-tick strips of the tape.

The chart may appear similar to the one shown in figure 5.2 below.

Figure 5.2

Note that:  The steps showing increasing velocity may vary slightly and become rather shorter at higher velocities. This is due to the effect of friction, particularly the drag of the tape through the ticker timer, which increases with speed.

To overcome these irregularities, draw a straight-line graph which passes through the centre of the tops of the tapes taking more notice of those which best fit a straight line through the origin.

Measure Dv and Dt from your tape or tape chart.

Since initial velocity, u = 0, Dv = Final velocity, v and Dt = time taken.

Calculate the acceleration due to gravity from the formula:

  Acceleration, g  = = ms^-2

How to calculate the value of g from given data.

(The steps may be understood full after covering ticker timer in Chapter 11).

Let:  Frequency of vibrator of the ticker timer = 50 Hz

 Number of two-tick tapes used to construct the chart = 10 pieces

The main steps in the calculations

Step I:  Calculate the time interval for the two-tick tape

Time interval of a two-tick  = Number of spaces x Tick

= 2 x

= 2 x

=

= 0.04 s

Step II:  Using the time interval for a two-tick tape and the number of pieces of tapes used calculate the time taken.

Time taken, t  = x

=   10 x 0.04

\ t  = 0.4 s

Step III:  Calculate the velocity, from the ticker tape.

Data:  Final velocity, v  =  = 14 cm in each 0.04s

  s = 14 cm = 0.14 m, t  = 0.04 s, v = ?

 Using the formula   v  =

=

= 3.5 ms^-1

Step IV: Calculate the acceleration due to gravity, g.

 Acceleration, g  = = ms^-1/s = 8.75 ms^-2

Note:  In the absence of all resisting forces such as the drag of the tape and air resistance,

the acceleration of freely falling object at the surface of the earth is about 9.8 ms^-2.

Self-Check  11.4

1.  A stone is dropped a deep hole. If it takes 10 seconds to hit the bottom of the hole.

Calculate:  (a)  its velocity.

(b)  the depth of the hole. (Take g = 10m/s²)

2.  A stone is dropped from a top of a cliff. If it takes 6 seconds to hit the ground,

calculate: (a)  the height of the cliff.

(b)  the velocity at impact. (Take g = 10 ms^-2)

3.  A tennis ball is thrown vertically upwards with an initial velocity of 40 ms^-1.

Calculate:  (a)  the maximum height reached.

 (b)  the time taken to reach the maximum height.

(Take g = 10 ms^-2 and neglect air resistance)

4.  A bullet is fired vertically upwards with an initial velocity of 300 ms^-1. Neglecting air resistance, calculate:

 (a)  the maximum height reached,

(b)  the time taken before it reaches the ground. (Take g = 10 ms^-2)

5.  A stone is thrown vertically upwards with an initial velocity of 20 ms^-1 from a tower 80 m high. Neglecting air resistance, calculate:

 (a)  the maximum height reached,

(b)  the time taken to reach the maximum height,

(c)  the total time which elapses before it just hits the ground

(Take g = 10 ms^-2)

11.6  Projectile Motion

A projectile is any body projected through space.  When two tennis balls placed at the same height are let to fall under gravity, one dropped vertically downwards and the other projected horizontally, it is noticed that all the balls hit the ground at the same time.

A diagram showing the motion of two tennis balls moving downwards from a table top (one dropped vertically downwards and the other projected horizontally)

h_v = vertical height, s = horizontal distance

 Facts about the motion

1. The initial velocity of the ball falling vertically is zero.
2. The vertical accelerations (due to gravity) of the two balls are equal.
3. A projectile falls like a body is dropped from rest.
4. The two balls take the same time to reach the ground.
5. The horizontal velocity of the projectile is independent of the vertical motion

(i.e. does not affect the vertical motion).

 The path of a projectile is called trajectory. It is (almost) the arc of a parabola from the following reasoning.

Consider a projectile with an initial velocity, u vertical velocity zero, initial horizontal acceleration zero and vertical acceleration downwards, g. After time, t, the horizontal distance, s, and vertical distance, h, traveled by the projectile are obtained by using the equation: s  = ut + ½gt²

For horizontal motion, g = 0  \ s  = ut  …………………………..  1

 For vertical motion, = 0   y  = ½gt²  ……………………………………….  2

Since x µ t  and y µ t², then:

y µ x²

 When this is plotted on a graph it is a parabola. That is the same as that of the graph in diagram for projectile motion but is inverted because the projectile falls and does not rise.

Worked Examples

1.  A ball is hit horizontally off a cliff with a velocity of 30 ms^-1. If it takes 5 seconds to reach the ground, calculate:

1. The height of the cliff.
2. The horizontal distance from the foot of the cliff to the point where the ball lands.
3. The velocity with which the ball hits the ground.

Solution:  u_y = 0 u_x = 30 ms^-1

(a)  To calculate the height of the cliff, we consider the vertical motion.

u_y = 0, g = 10 ms^-2, t = 5 s, s = h = ?

Using the formula s = u_yt +½gt²

= (0x5) + ½ x 10 x 5²

= 0 + 5 x 25

s = h  = 125 m

(b)  The horizontal distance, s, is got by considering the horizontal motion.

 u_x = 30, g = 0 ms^-2, t = 5 s, s = ?

 Using the formula s = u_xt +½gt² = 30 x 5 + ½ x 0 x 5² = 150 + 0 = 150 m

Therefore, the ball lands 150 m from the foot of the cliff.

(c)  The velocity with which the ball hits the ground is got by considering the vertical motion.

This is because the horizontal velocity of the ball remains unchanged throughout the motion.

Vertical initial velocity, u_y = 0, Vertical final velocity, v_y = ?,

g = 10 ms^-2, t = 5 s,

Using the formula: v  = u + at

  v_y  = u_y + at

= 0 + 10 x 5

= 50 ms^-1

1. A bomb is released from a jet fighter plane moving with velocity of 400 ms^-1 to hit a rebel camp in northern Uganda. If the bomb took 10 seconds to hit the target, calculate:
   1. The altitude at which the bomb was released.
   2. The horizontal distance from the vertical point of the plane to the target.
   3. The velocity with which the bomb hits the target. (Take g = 10 ms^-2)

Solution  u_y = 0 ms^-1, g = 10 ms^-2, t = 10 s, s = h = ?

(a)  To calculate the altitude of the plane, we consider the vertical motion.

Using the formula: s  = u_yt +gt²

= (0x5) +x 10 x 10²

= 0 + 5 x 100

  s = h  = 500 m

(b)  The horizontal distance, s, is got by considering the horizontal motion.

u_x = 400, g = 0 ms^-2, t = 10 s, s = ?

s  = u_xt + ½gt²

= 400 x 10 + ½ x 0 x 10²

= 4000 + 0

s  = 4000 m

(c)  The horizontal velocity of the bomb remains unchanged throughout the motion

(400 ms^-1). The vertical velocity increases as the bomb falls.

 The velocity with which the bomb hits the ground is got by considering the vertical motion.

u_y = 0 ms^-1, g = 10 ms^-2, t = 10 s,

v  = u + at

 = u_y + gt

 = 0 + 10 x 10

 = 100 ms^-1

Self-Check  11.5

1.  Which one of the following statements is true when a stone of mass 2 kg and that of 1 kg are released from the same point at the same time?

A. both masses will hit the ground at the same time.  

B. the 2kg mass will hit the ground first.

C. the 1 kg mass will hit the ground first.

D. they fall with different speeds.

2.  A mass is projected upwards with a velocity of 10 m/s. If the acceleration due to gravity is 10 m/s ², what is the maximum height reached in meters?

A. 1 B. 5 C. 10 D. 20 E. 100

3.  A stone is hit horizontally off a cliff with a velocity of 50 ms^-1. If it takes 10 seconds to reach the ground, calculate:

1. The height of the cliff.
2. The horizontal distance from the foot of the cliff to the point where the ball lands.
3. The velocity with which the ball hits the ground.

4.  An Israel jet fighter plane moving with velocity of 300 ms^-1 released a bomb to hit a Hezbollah base in Southern Lebanon. If the bomb took 20 seconds to hit the target, calculate:

(a)  The altitude at which the bomb was released.

(b)  The horizontal distance from the vertical point of the plane.

(c)  The velocity with which the bomb hits the target. (Take g = 10 ms^-2)

5.  A stone falls from rest from the top of a high tower. Ignoring air resistance and taking g =10m/s², what is the velocity and distance travelled after:

(i)  1 s,  

(ii)  2s,  

(iii)  3s  

(iv)  5s?

11.7  Circular Motion

 One of the effects of force is that it keeps bodies moving in a circular path about a fixed point. This kind of motion is described as Circular Motion.

Examples of circular motion include:

• a stone/pendulum bob whirled at the end of a string in a vertical or horizontal plane,
• a vehicle negotiating a corner,
• an electron orbiting a nucleus,
• a planet or satellite orbiting the earth.

 Facts about a body moving in a circle:

1. Its direction changes every second.
2. Its velocity is constantly changing due to the change in direction of the motion.
3. It has an acceleration called centripetal acceleration due to the constant change in velocity.
4. It has a force called Centripetal force, acting on it producing the acceleration.
5. The centripetal force is perpendicular to the direction of the motion.
6. Both the centripetal force and the centripetal acceleration are acting towards the centre of the circle.
   

    

Experimental results show that the centripetal force required to keep a body in a circular path increases with:

(i)  an increase in the mass of the body.

(ii)  an increase in the speed of the body and

(iii)  a decrease in the radius of the circular path.

When the object is released, it moves in a straight line. The direction of the motion is along the tangent to the circle as shown in the diagram below.

A diagram showing a bob tied at the end of a string being whirled in a horizontal circle

Self-Check  11.6

1.  (a)  Define the term acceleration.

(b)  A body attached to a string is swung in a vertical circular path in air as shown in the figure below.

Copy the above diagram and on it indicate and name all the forces acting on the body if the body is moving in an anti-clockwise direction.

1. Explain why the weight of an object on the earth’s surface may vary from one place to another.