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Kinematic Equations (HL IB Physics)

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Kinematic Equations

  • The kinematic equations of motion are a set of four equations that can describe any object moving with constant or uniform acceleration
  • They relate the five variables:
    • s = displacement
    • u = initial velocity
    • v = final velocity
    • a = acceleration
    • t = time interval

Kinematic Equations of Motion

  • There are four kinematic equations:

Summary of equations, downloadable AS & A Level Physics revision notes

How to derive the kinematic equations

Deriving v=u+at, downloadable AS & A Level Physics revision notes

Derivation of v = u + at

Deriving s=(u+v)t 2, downloadable AS & A Level Physics revision notes

Deriving s=ut+12at2, downloadable AS & A Level Physics revision notes

Deriving v2=u2+2as, downloadable AS & A Level Physics revision notes

This final equation can be derived from two of the others

Key Takeaways

  • The key terms to look out for are:

'Starts from rest'

  • This means u = 0 and t = 0
  • This can also be assumed if the initial velocity is not mentioned

'Falling due to gravity'

  • This means a = g = 9.81 m s–2 
    • It doesn't matter which way is positive or negative for the scenario, as long as it is consistent for all the vector quantities
    • If downwards is considered positive, this is 9.81 m s–2 , otherwise, it is –9.81 m s–2 
  • For example, if downwards is negative then for a ball travelling upwards, s must be positive and a must be negative

'Constant acceleration in a straight line'

  • This is a key indication for the kinematic equations are intended to be used
    • For example, an object falling in a uniform gravitational field without air resistance

How to use the kinematic formulae

  • Step 1: Write out the variables that are given in the question, both known and unknown, and use the context of the question to deduce any quantities that aren’t explicitly given
    • e.g. for vertical motion a = ± 9.81 m s–2, an object which starts or finishes at rest will have u = 0 or v = 0

  • Step 2: Choose the equation which contains the quantities you have listed
    • e.g. the equation that links s, u, a and t is s = ut + ½at2

  • Step 3: Convert any units to SI units and then insert the quantities into the equation and rearrange algebraically to determine the answer

Worked example

The diagram shows an arrangement to stop trains that are travelling too fast.

SUVAT Problems Worked Example (1)

 At marker 1, the driver must apply the brakes so that the train decelerates uniformly in order to pass marker 2 at no more than 10 m s–1.

The train carries a detector that notes the times when the train passes each marker and will apply an emergency brake if the time between passing marker 1 and marker 2 is less than 20 s.

Trains coming from the left travel at a speed of 50 m s–1.

Determine how far marker 1 should be placed from marker 2.

Answer:

SUVAT Problems Worked Example (2), downloadable AS & A Level Physics revision notes

Worked example

A cyclist is travelling directly east through a village, which is completely flat, at a velocity of 6 m s–1. They then start to constantly accelerate at 2 m s–2 for 4 seconds.

(a)
Calculate the distance that the cyclist covers in the 4 s acceleration period.
(b)
Calculate the cyclist's final velocity after the 4 s interval of acceleration.
 

Later on in their journey, cyclist A is now cycling through a different village, at a constant velocity of 18 m s–1. Cyclist A passes a friend, Cyclist B who begins accelerating from rest at a constant acceleration of 1.5 m s–2 in the same direction as Cyclist A at the moment they pass.

(c)
Calculate how long it takes for Cyclist B to catch up to Cyclist A
 

Answer:

(a) Calculate the displacement, s

Step 1: List the known quantities

  • Initial velocity, u = 6 m s–1
  • Acceleration, a = 2 m s–2 
  • Time, t = 4 s
  • Displacement = s (this needs to be calculated)

Step 2: Identify the best SUVAT equation to use

  • Since the question states constant acceleration, the kinematic equations can be used
  • In this problem, the equation that links s, u, a, and t is

s space equals space u t space plus space 1 half a t squared

Step 3: Substitute the known quantities into the equation

s = (6 × 4) + (0.5 × 2 × 42) = 24 + 16

Displacement:  s = 40 m

(b) Calculate the final velocity, v

Step 1: List the known quantities

  • Initial velocity, u = 6 m s–1
  • Acceleration, a = 2 m s–2
  • Time, t = 4 s
  • Final velocity = v (this needs to be calculated)

Step 2: Identify and write down the equation to use

  • Since the question states constant acceleration - SUVAT equation(s) - can be used
  • In this problem, the equation that links v, u, a, and t is:

v space equals space u space plus space a t

Step 3: Substitute the known quantities into the equation

v = 6 + (2 × 4)

Final velocity:  v = 14 m s–1

(c) Calculate the time t for B to catch up to A

Step 1: List the known quantities for cyclist A

  • Initial velocity, u subscript A = 18 m s–1
  • Acceleration, a subscript A = 0 m s–2
  • Time = t
  • Displacement = s subscript A

Step 2: List the known quantities for cyclist B

  • Initial velocity, u = 0 m s–1
  • Acceleration, a = 1.5 m s–2 
  • Time = t
  • Displacement = s subscript B

Step 3: Write expressions for Cyclist A and Cyclist B in terms of their displacement

  • Cyclist A's motion can be expressed by:

s subscript A space equals space u subscript A t space plus space 1 half a subscript A t squared

s subscript A space equals space 18 t space plus space 0 space equals space 18 t

  • Cyclist B's motion can be expressed by:

s subscript B space equals space u subscript B t space plus space 1 half a subscript B t squared

s subscript B space equals space 0 space plus space open parentheses 1 half cross times 1.5 cross times t squared close parentheses space equals space 3 over 4 t squared

Step 4: Equate the two equations and solve for t

  • The two equations describe the displacement of each cyclist respectively
  • When equating them, this will find the time when the cyclists are at the same location

s subscript A space equals space s subscript B

18 t space equals space 3 over 4 t squared space space space space space rightwards double arrow space space space space space 3 over 4 t squared space minus space 18 t space equals space 0

open parentheses t squared space minus space 24 t close parentheses space equals space 0

  • Therefore, solving for t, it can be two possible answers:

t = 0 s or t = 24 s

  • Since the question is seeking the time when the two cyclists meet after first passing each other, the final answer is 24 s

Worked example

A science museum designed an experiment to show the fall of a feather in a vertical glass vacuum tube.

The time of fall from rest is 0.5 s.

WE - Projectile Motion Worked Example 1 question image, downloadable AS & A Level Physics revision notes

Use an appropriate SUVAT equation to calculate the length L of the tube.

Answer:

WE - Projectile Motion Worked Example 1 answer image, downloadable AS & A Level Physics revision notes

Exam Tip

This is one of the most important sections of this topic - usually, there will be one, or more, questions in the exam about solving problems with the kinematic equations equations. The best way to master this section is to practice as many questions as possible!

Watch out for the direction of vectors: displacement, acceleration and velocity. Take a single direction as positive (and hence the opposite direction is negative) and stick with it throughout the question, this is the most common pitfall.

Don't worry, you won't have to memorise these, they are give in your data booklet in the exam.

You may sometimes see these equations referred to as 'SUVAT' equations.

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Ashika

Author: Ashika

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.