When the acceleration is constant, five quantities describe the motion: \(s, u, v, a, t\). Know any three and one of five equations gives you the rest.
For constant acceleration in a straight line: \(v = u + at\), \(s = \tfrac{1}{2}(u + v)t\), \(s = ut + \tfrac{1}{2}at^2\), \(s = vt - \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\). Each one leaves out one of the five quantities, so pick the equation without the one you neither know nor want.
Fig. 1: with constant acceleration the velocity-time graph is a straight line. Gradient a gives v = u + at; the area ut + ½(v − u)t = ½(u + v)t is the displacement
Section 01
Where the equations come from.
In the lab, release the ball from a chosen height and step through the strobe images. The gaps grow evenly because the acceleration is constant, which is exactly when the suvat equations apply.
Equation
Leaves out
\(v = u + at\)
\(s\)
\(s = \tfrac{1}{2}(u + v)t\)
\(a\)
\(s = ut + \tfrac{1}{2}at^2\)
\(v\)
\(s = vt - \tfrac{1}{2}at^2\)
\(u\)
\(v^2 = u^2 + 2as\)
\(t\)
From Fig. 1: the gradient gives \(v = u + at\) and the trapezium area gives \(s = \tfrac{1}{2}(u + v)t\). The other three follow by eliminating \(v\), \(u\) or \(t\).
The lab uses \(g = 9.81\); in Paper 4 use \(g = 10\ \text{m s}^{-2}\).
Worked example Choosing the equation
A car travelling at \(25\ \text{m s}^{-1}\) brakes uniformly and stops in 60 m. Find the deceleration and the time taken.
Known: \(u = 25\), \(v = 0\), \(s = 60\). Want \(a\), not \(t\): use \(v^2 = u^2 + 2as\).
For \(t\), avoid the rounded \(a\): \(s = \tfrac{1}{2}(u + v)t\) gives \(60 = 12.5t\).
Answer: Deceleration \(5.21\ \text{m s}^{-2}\); time 4.8 s.
Worked example Distance in a particular second
A particle starts with velocity \(3\ \text{m s}^{-1}\) and accelerates at \(2\ \text{m s}^{-2}\). How far does it travel in the 5th second?
The 5th second runs from \(t = 4\) to \(t = 5\).
\(s(5) = 3(5) + \tfrac{1}{2}(2)(25) = 40\) m; \(s(4) = 3(4) + \tfrac{1}{2}(2)(16) = 28\) m.
Distance in the 5th second \(= 40 - 28\).
Answer: 12 m
Examiner trap
The suvat equations need CONSTANT acceleration. If a question gives \(a\) or \(v\) as a function of \(t\), use calculus instead (next lesson).
Section 02
Motion under gravity.
A particle moving freely up and down has acceleration \(g = 10\ \text{m s}^{-2}\) downwards, whether it is rising, at the top, or falling. Choose up as positive and write \(a = -10\).
Worked example Thrown up from a height
A ball is thrown vertically upwards at \(14\ \text{m s}^{-1}\) from a point 2 m above the ground. Find the greatest height above the point of projection and the time it takes to reach the ground.
Up positive, \(a = -10\). At the top \(v = 0\): \(0 = 196 - 20s\), so \(s = 9.8\) m.
The ground is 2 m BELOW the start, so \(s = -2\): \(-2 = 14t - 5t^2\).
\(5t^2 - 14t - 2 = 0\), so \(t = \dfrac{14 + \sqrt{236}}{10}\) (the other root is negative).
\(t = 2.94\) s.
Answer: 9.8 m above the point of projection; 2.94 s.
Your turn
A stone is dropped from rest off a cliff and hits the sea 3.2 s later. Find the height of the cliff and the speed of impact.
Using \(s = +2\) for the drop to the ground. Displacement is measured from the point of projection, so landing below it means a NEGATIVE displacement when up is positive.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which suvat equation does not involve \(v\)?
Quick check+10 XP
A particle has \(u = 4\ \text{m s}^{-1}\) and \(a = 3\ \text{m s}^{-2}\). What is its velocity after 5 s?
Quick check+10 XP
An object is dropped from rest. How far does it fall in the first 2 s? (\(g = 10\))
Quick check+10 XP
A particle moving at \(12\ \text{m s}^{-1}\) decelerates at \(3\ \text{m s}^{-2}\). How long does it take to stop?
Section 03
Two stages and two particles.
When the acceleration changes, treat each stage separately: the final velocity of one stage is the initial velocity of the next. When two particles move, write each one's position at time \(t\) from the same origin and set them equal.
Worked example Meeting in mid-air
Particle A is released from rest at the top of a 45 m tower. At the same instant particle B is projected vertically upwards at \(20\ \text{m s}^{-1}\) from the ground directly below. Find when and at what height they meet.
Height of A: \(h_A = 45 - 5t^2\). Height of B: \(h_B = 20t - 5t^2\).
Meet when \(45 - 5t^2 = 20t - 5t^2\): the \(t^2\) terms cancel, so \(t = 2.25\) s.
Height \(= 20(2.25) - 5(2.25)^2 = 19.7\) m.
(B's velocity then is \(20 - 22.5 = -2.5\ \text{m s}^{-1}\): it is already falling.)
Answer: After 2.25 s, at 19.7 m above the ground.
Worked example Find the unknowns from two facts
A car passes point A with speed \(u\) and accelerates uniformly, passing B, 100 m further on, 5 s later at \(30\ \text{m s}^{-1}\). Find \(u\) and the acceleration.
Restart the clock carefully. If B is released 1 s after A, then B has been moving for \((t - 1)\) s when A has been moving for \(t\) s.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
A ball is thrown vertically upwards at \(14\ \text{m s}^{-1}\) from a point 2 m above the ground. Find the time it takes to reach the ground.
Exam style+20 XP
A particle starts with velocity \(3\ \text{m s}^{-1}\) and constant acceleration \(2\ \text{m s}^{-2}\). Find the distance travelled in the 5th second.
Exam style+20 XP
A car passes A and accelerates uniformly, passing B, 100 m further on, 5 s later with speed \(30\ \text{m s}^{-1}\). Find the acceleration.
Exam style+20 XP
A is released from rest at the top of a 45 m tower; at the same instant B is projected upwards at \(20\ \text{m s}^{-1}\) from the ground directly below. At what height do they meet?
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