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Equations of motion for OCR GCSE Physics

Equations of motion

This page covers the following topics:

1. Basic equation of motion
2. Basic equations for uniform acceleration

For an object moving at a constant velocity, the distance it has travelled during a given time interval at a given speed can be calculated using the following formula: distance = speed ร— time. The instantaneous velocity of an object is its velocity at a specific point in time during its motion, whereas its average velocity is the average of all instantaneous velocities during the object's motion.

Basic equation of motion

The initial velocity, final velocity, acceleration and displacement of an object travelling at a uniform acceleration can be calculated using the following formula: vยฒ โˆ’ uยฒ = 2as.

Basic equations for uniform acceleration

1

A ball is dropped from a certain height above the ground. Given that gravitational acceleration is 9.8 m/sยฒ and that the ball lands on the ground at 7 m/s, calculate the height from which the ball is dropped.

Since the ball is dropped, u = 0.
vยฒ โˆ’ uยฒ = 2as
7ยฒ โˆ’ 0ยฒ = 2 ร— 9.8 ร— s
s = 49 รท 19.6 = 2.5 m

2.5 m

A ball is dropped from a certain height above the ground. Given that gravitational acceleration is 9.8 m/sยฒ and that the ball lands on the ground at 7 m/s, calculate the height from which the ball is dropped.

2

How long must the motion of an object travelling at a constant speed of 14 m/s be to cover 65 m?

distance = speed ร— time
t = 65 รท 14 = 4.64 s (3 s. f.)

4.64 s

How long must the motion of an object travelling at a constant speed of 14 m/s be to cover 65 m?

3

Define instantaneous velocity.

The instantaneous velocity of an object is its velocity at a specific point in time.

velocity at a specific time

Define instantaneous velocity.

4

An object covers 25 m in 8 seconds. Calculate the average speed of the object.

Using d = s ร— t,
average speed = 25 m/ 8 s = 3.125 m/s.

3.125 m/s

An object covers 25 m in 8 seconds. Calculate the average speed of the object.

5

A runner accelerates from 2.2 m/s to 3.4 m/s. Given that her acceleration is 1.3 m/sยฒ, calculate the distance she covers during her acceleration to 2 decimal places.

vยฒ โˆ’ uยฒ = 2as
3.4ยฒ โˆ’ 2.2ยฒ = 2 ร— 1.3 ร— s
s = 6.72 รท 2.6 = 2.58 m

2.58 m

A runner accelerates from 2.2 m/s to 3.4 m/s. Given that her acceleration is 1.3 m/sยฒ, calculate the distance she covers during her acceleration to 2 decimal places.

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