Which equation of motion does not contain time?

v² = u² + 2as. Use it when the time is neither given nor wanted.

When can the equations of motion be used?

only when the acceleration is constant. They are derived assuming uniform acceleration.

A ball is thrown straight up. What is its acceleration at the highest point?

9.81 m/s² downwards. Gravity acts on it all the time.

Equations of motion · Edexcel IAL Physics WPH revision

In plain words

When something moves with a steady acceleration, four equations link how far it goes, how fast it starts and ends, how quickly it speeds up, and how long it takes. Know any three of those five and you can find the other two.

Three things to know

v = u + at. s = ut + ½at². v² = u² + 2as. s = ½(u + v)t. They only apply when the acceleration is constant.
Choose a positive direction and stick to it. Anything pointing the other way is negative.
In free fall with no air resistance, the acceleration is g = 9.81 m/s² downwards, whether the object is going up, coming down or at the top.

Worked example

A car accelerates uniformly from 10 m/s to 30 m/s over 200 m. Find its acceleration.

List what you know: u = 10, v = 30, s = 200. You want a, and there's no t.
The equation without t is v² = u² + 2as: 900 = 100 + 400a.
a = 2.0 m/s².

Tips and tricks

Write down s, u, v, a, t with what you know. Pick the equation that has your three knowns and your one unknown.
At the top of its flight a thrown ball has zero velocity but its acceleration is still g.

5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Equations of motion: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

Which equation of motion does not contain time?
- v = u + at
- s = ut + ½at²
- v² = u² + 2as (the answer)
- s = ½(u + v)t
Use it when the time is neither given nor wanted.
An object starts from rest and accelerates at 2.0 m/s² for 5.0 s. How far does it travel?
- 10 m
- 25 m (the answer)
- 50 m
- 100 m
½ × 2.0 × 5.0².
A stone falls from rest. What is its speed after 3.0 s? (Ignore air resistance.)
- 9.81 m/s
- 19.6 m/s
- 29.4 m/s (the answer)
- 44.1 m/s
9.81 × 3.0.
When can the equations of motion be used?
- always
- only when the acceleration is constant (the answer)
- only when the velocity is constant
- only for falling objects
They are derived assuming uniform acceleration.
A ball is thrown straight up. What is its acceleration at the highest point?
- zero
- 9.81 m/s² downwards (the answer)
- 9.81 m/s² upwards
- it depends on the mass
Gravity acts on it all the time.

5 questions

Tap an answer and you’ll see straight away whether it’s right, and why.

Worksheet

4 questions, 10 marks. Write your answers on paper, then check them.

Equations of motion

Edexcel International A Level Physics WPH · 10 marks · papermunch.org

Name ______________________________ Date ______________

A ball is dropped from rest and falls 20 m. Find the time taken and its speed on landing.[3]
Show answerHide answer
2.02 s and 19.8 m/s. t = √(2 × 20 ÷ 9.81); v = 9.81 × 2.02.
I got this one
A stone is thrown vertically upwards at 15 m/s. Find the maximum height it reaches.[2]
Show answerHide answer
11.5 m. 15² ÷ (2 × 9.81).
I got this one
A train slows uniformly from 25 m/s to rest in 50 s. Find the distance it travels.[2]
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625 m. ½ × (25 + 0) × 50.
I got this one
Describe how to measure the acceleration of free fall using a falling object.[3]
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Drop a small, dense ball from rest through a measured height h, and time the fall with light gates or an electromagnet and trapdoor connected to a timer. Repeat for several heights. Since h = ½gt², a graph of h against t² is a straight line through the origin with gradient ½g.
I got this one

Answers: Equations of motion

1. 2.02 s and 19.8 m/s. t = √(2 × 20 ÷ 9.81); v = 9.81 × 2.02.
2. 11.5 m. 15² ÷ (2 × 9.81).
3. 625 m. ½ × (25 + 0) × 50.
4. Drop a small, dense ball from rest through a measured height h, and time the fall with light gates or an electromagnet and trapdoor connected to a timer. Repeat for several heights. Since h = ½gt², a graph of h against t² is a straight line through the origin with gradient ½g.