The four equations of uniformly accelerated motion connect five quantities: s, u, v, a and t. Every question gives three and asks for one, so exactly one of the four equations leaves out the quantity that you do not need.
This lesson is part of SPM Physics force and motion I. It builds on interpreting displacement, velocity and acceleration.
How do I choose the equation?
Use a short list-and-cross-out routine.
- Write s, u, v, a and t in a column.
- Fill in the three values that the question gives, including hidden ones such as u = 0 for “from rest”.
- Mark the quantity you want.
- Choose the equation that does not contain the remaining unknown.
| Equation | Missing quantity |
|---|---|
| v = u + at | s |
| s = ut + ½at² | v |
| v² = u² + 2as | t |
| s = ½(u + v)t | a |
Worked example 1: no time given
A cyclist speeds up from 6 m s⁻¹ to 14 m s⁻¹ over 100 m. Find the acceleration and the time taken.
Given: u = 6, v = 14, s = 100. Unknown: a. The time is missing, so use v² = u² + 2as.
196 = 36 + 2a(100), so 160 = 200a and a = 0.8 m s⁻². Now find t from v = u + at: 14 = 6 + 0.8t, so t = 10 s. Check with s = ½(u + v)t = ½ × 20 × 10 = 100 m, which matches.
Worked example 2: a negative acceleration
A car travelling at 20 m s⁻¹ brakes with a constant deceleration of 5 m s⁻². Find the stopping distance.
Take forward as positive, so a = −5 m s⁻². The car stops, so v = 0. Given: u = 20, v = 0, a = −5. Unknown: s, with time missing, so use v² = u² + 2as.
0 = 400 + 2(−5)s, so 10s = 400 and s = 40 m. If you forget the minus sign on a, you get s = −40 m, and the negative distance tells you to recheck the sign.
The mistake to avoid
The common mistake is to choose an equation first and hunt for numbers afterwards. That leads to an equation with two unknowns, and the student then guesses a value.
| Habit | Result |
|---|---|
| Pick the equation first | Two unknowns, stuck or guess |
| List s, u, v, a, t first | The equation follows from the missing quantity |
Also watch the units. A speed in km h⁻¹ must be converted to m s⁻¹ by dividing by 3.6 before you substitute. The units and significant figure checker helps you test a conversion.
Check yourself
A ball is thrown straight up at 15 m s⁻¹. Taking g = 10 m s⁻², find the maximum height and the time to reach it.
Answer
Take upward as positive, so a = −10 m s⁻². At the top, v = 0. Given: u = 15, v = 0, a = −10.
Height: v² = u² + 2as gives 0 = 225 − 20s, so s = 11.25 m. Time: v = u + at gives 0 = 15 − 10t, so t = 1.5 s.
What to study next
Go on to explaining momentum and impulse, where the same quantities appear in collisions. You can test the skill in the force and motion I practice set. For a method that checks the answer, see selecting and checking a physical model.
If you want a teacher to watch your choices with you, see online one-to-one Physics tuition.