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:
How to derive the kinematic equations
Derivation of v = u + at
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 = 9.8 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.
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:
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.
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.
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
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:
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, = 18 m s–1
- Acceleration, = 0 m s–2
- Time = t
- Displacement =
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 =
Step 3: Write expressions for Cyclist A and Cyclist B in terms of their displacement
- Cyclist A's motion can be expressed by:
- Cyclist B's motion can be expressed by:
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
- 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.
Use an appropriate SUVAT equation to calculate the length L of the tube.
Answer:
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.