IGCSE Physics Motion And Energy

2. Speed – Distance travelled by any object in unit time is known as speed.
Speed – Distance travelled (S.I. unit = m/s)
Time
Avg. Speed – Initial speed + Final speed (where acceleration is constant)
2
For e.g., -
If a man travels 2 km in 20 mins, then he catches a train and covers
another 20 km in 15 mins. Find the avg. speed in its S.I. unit ?
Solution –
Speed= total distance travelled
Time
= (2000 + 20000) 1km = 1000m
2100 1min = 60sec
= (22000)
2100
= 10.48 m/s

3. Velocity – It is the distance travelled in unit time in a straight direction
or simply speed in a particular direction.
Velocity (v) = Distance move in straight direction (SI
Unit=m/s)
Time Taken
Distance travelled on a straight line is known as displacement.
Therefore, v = Displacement
Time Taken
Acceleration – Acceleration is the change in velocity in unit time.
Acceleration(a)= Change in velocity (SI
Unit=m/s2)
Time taken to change
So,
a= v - u (v = final velocity, u = initial velocity, t =

5. Distance – Time Graph
Slope of distance time graph gives speed.
If the line is straight the object is travelling with uniform speed or velocity.
If the line is parallel to the x-axis the object is at rest.

6. • Slope of velocity time graph shows acceleration
• If line is straight and is making a angle with the x axis
there is uniform acceleration
• If the line is parallel to the x axis the object is in
constant velocity
• Area enclosed inside the graph shows the distance

7. Problems involving uniform acceleration can be solved
quickly using equation of motion.
First equation v = final velocity
v = u + at u = initial velocity
S = distance
Second equation t = time
S = ut + ½ at2 a = acceleration
Third equations
v2 = u2 + 2aS

8. Speed
5
2.5
2 4
6 8 10 12 14
Time(m/s)
The graph shows a incomplete velocity time graph for a boy running 100 m
Find the acceleration during the first 4 seconds
a = (v – u) / t = (5 – 0)/4 = 5/4 m/s2 =1.25 m/s2
How far does the boy travel in 4 seconds
Distance travelled can be found out by
a) equation of acceleration
b) by area enclosed in side graph
s = [(u + v)/2]t = 2.5 x 4= 10m
Other method
Area of triangle=1/2x base x height
½ x 4 x 5 = 10m

9. First Law
The body stays at rest, or if moving it continues to move with uniform
velocity, unless an external force makes it behave differently.
For Example – A ball rolling on the ground will never come to a stop
if the friction does not act on it in a an opposite direction.
Mass And Inertia
The tendency of the object to resist any change in its state of rest or
motion is known as inertia. Its effect is seen when a car stops
suddenly and the occupants lurch forward in an attempt to continue
moving. The larger the mass of the body the larger the inertia
therefore we say mass is the measure of inertia.
For example - Its difficult to move Priyansh than to move Gaurav.

10. Second Law
The rate of change of momentum is directly proportional to the external
force acting on the body.
Which gives the result as:-
F=ma
one newton is defined as the force which gives a mass of 1 kg, an acceleration
of 1 m/s2.
Weight And Gravity
The Weight W of a body is the force of gravity acting on it which gives it
acceleration g when it is falling.
If a body has mass m then,
F = ma
W = mg ( force = weight, g = acceleration due to gravity)
Where g = 9.8 N/kg = 9.8 m/s2.
Third Law
If a body A exerts a force on body B, then body B exerts an equal but
opposite force on body A.
Newton’s 1st law

11. Energy is the ability to do work.

12. Kinetic And Potential Energy
The kinetic energy of an object is the energy which it possesses due to its motion.
Ek = ½ x mv2
Where Ek = kinetic energy, m = mass, v = velocity
The potential energy of a body is the energy it has because of its position or
configuration. It is having a very wide meaning as p.e. is of different types for e.g.
gravitational p.e., electrostatic p.e., magneto static p.e., chemical p.e. Here while
using term potential energy we mean by gravitational p.e. only.
Ep = mgh
Where m = mass, g = 9.8 m/s2, h = given height
A mass m at height h above the ground has p.e. = mgh, when it falls, its velocity
increases and it gains k.e. at the expense of its p.e as followed by law 0f
conservation of energy. If it starts from rest and air resistance is negligible its k.e.
just before touching the ground equals the p.e. lost by it.
Thus,
½ x mv2 = mgh

13. Question: A soccer ball of mass 450 g is travelling at a speed of 20m/s. How
much kinetic energy does the soccer ball have?
Answer
Step 1 : Identify the quantities involved
The quantities involved are the mass of the soccer ball (450 g). The speed
of the soccer ball (20 m/s) and KE the kinetic energy transferred.
Step 2 : Identify the equation needed
Here we where asked to find the kinetic energy so we use the equation
KE = ½ x mv2. It is important to note that in order to apply this equation
all quantities must be in SI units.
Step 3 : Convert to SI units
The mass of the soccer ball is in g which is not a SI. Converting we have 450 g
is 0.45 kg. The speed of the ball is already in SI units.
Step 4 : Determine solution
Inserting the quantities into the equation we get
KE = ½ x mv2
= ½ 0.45kg (20m/s)2
= 90J

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21. Made By -
Ayush & Dhairya