Constant Acceleration (suvat) – A Level Maths Year 1

1 The Five Equations

When the acceleration is constant, motion in a straight line is described by five variables and five equations (DfE content area Q · Kinematics). They are known as the suvat equations after the five letters.

Symbol	Meaning	Units
$s$	displacement from the starting point	m
$u$	initial velocity	m s$^{-1}$
$v$	final velocity	m s$^{-1}$
$a$	(constant) acceleration	m s$^{-2}$
$t$	time taken	s

The suvat equations (each omits one variable)

$v = u + at$   (no $s$)
$s = \dfrac{(u + v)}{2}\,t$   (no $a$)
$s = ut + \tfrac12 at^2$   (no $v$)
$s = vt - \tfrac12 at^2$   (no $u$)
$v^2 = u^2 + 2as$   (no $t$)

The method: list the five letters, fill in the three you know, mark the one you want, and choose the equation that leaves out the fifth.

Only for constant acceleration. If the acceleration depends on time — for example $a = 6t - 2$ — the suvat equations are wrong, and you must use calculus (see Variable Acceleration).

2 Deriving the Equations from a Velocity–Time Graph

You may be asked to derive the equations. With constant acceleration the $v$–$t$ graph is a straight line from $(0,\,u)$ to $(t,\,v)$.

Worked Example 1 — The derivation

①Gradient $= a$, so $\dfrac{v - u}{t} = a$, giving $\boxed{v = u + at}$.

②Area of the trapezium $= s$, so $\boxed{s = \tfrac12(u + v)t}$.

③Substitute $v = u + at$ into ②: $s = \tfrac12(2u + at)t = \boxed{ut + \tfrac12 at^2}$ — the rectangle plus the triangle.

④Substitute $u = v - at$ into ②: $s = \tfrac12(2v - at)t = \boxed{vt - \tfrac12 at^2}$.

⑤From ①, $t = \dfrac{v - u}{a}$. Into ②: $s = \dfrac{(u + v)(v - u)}{2a} = \dfrac{v^2 - u^2}{2a}$, so $\boxed{v^2 = u^2 + 2as}$.

3 Using the Equations

Worked Example 2 — Braking distance

A car travelling at $20$ m s$^{-1}$ brakes with constant deceleration $5$ m s$^{-2}$. Find the stopping distance and stopping time.

①$u = 20$, $v = 0$, $a = -5$; want $s$ and $t$.

②$v^2 = u^2 + 2as$: $0 = 400 - 10s \Rightarrow s = 40$ m.

③$v = u + at$: $0 = 20 - 5t \Rightarrow t = 4$ s.

Worked Example 3 — Finding $u$ and $a$

A particle accelerating uniformly travels $30$ m in $4$ s, finishing with speed $10$ m s$^{-1}$. Find its initial speed and its acceleration.

①$s = 30$, $t = 4$, $v = 10$.

②$s = \tfrac12(u + v)t$: $30 = \tfrac12(u + 10)(4) = 2(u + 10) \Rightarrow u = 5$ m s$^{-1}$.

③$v = u + at$: $10 = 5 + 4a \Rightarrow a = 1.25$ m s$^{-2}$.

Worked Example 4 — Returning to the start

A particle passes $O$ at $5$ m s$^{-1}$ with constant acceleration $-0.5$ m s$^{-2}$. Find the greatest distance it reaches from $O$ (in the positive direction) and when it returns to $O$.

①Greatest distance when $v = 0$: $0 = 25 + 2(-0.5)s \Rightarrow s = 25$ m.

②Back at $O$ when $s = 0$: $0 = 5t - 0.25t^2 = t(5 - 0.25t)$.

③$t = 0$ (the start) or $t = 20$ s.

One suvat equation covers the whole motion — out and back — because the acceleration is the same throughout. You do not need to split it at the turning point.

4 Vertical Motion under Gravity

Ignoring air resistance, a particle moving vertically has constant acceleration $g = 9.8$ m s$^{-2}$ downwards, whether it is going up, coming down or at the top. Choose a positive direction (usually upwards, so $a = -9.8$) and keep it for the whole problem.

Situation	What you know
Dropped / released from rest	$u = 0$
Greatest height	$v = 0$
Returns to the point of projection	$s = 0$
Lands below its start (e.g. sea below a cliff)	$s$ is negative if up is positive

Worked Example 5 — Thrown up from a cliff

A stone is thrown vertically upwards at $12$ m s$^{-1}$ from the top of a cliff $40$ m above the sea. Find the time it takes to reach the sea and its speed as it hits the water.

①Up positive: $u = 12$, $a = -9.8$, $s = -40$.

②$s = ut + \tfrac12 at^2$: $-40 = 12t - 4.9t^2 \Rightarrow 4.9t^2 - 12t - 40 = 0$.

③$t = \dfrac{12 \pm \sqrt{144 + 784}}{9.8} = \dfrac{12 \pm \sqrt{928}}{9.8}$; the positive root is $t = 4.33$ s (3 s.f.).

④$v^2 = u^2 + 2as = 144 + 2(-9.8)(-40) = 928 \Rightarrow$ speed $= \sqrt{928} = 30.5$ m s$^{-1}$ (3 s.f.).

The negative root $t = -1.88$ is rejected — it is a time before the stone was thrown.

Worked Example 6 — Time above a height

A ball is projected vertically upwards at $20$ m s$^{-1}$ from ground level. For how long is it at least $15$ m above the ground?

①$15 = 20t - 4.9t^2 \Rightarrow 4.9t^2 - 20t + 15 = 0$.

②$t = \dfrac{20 \pm \sqrt{400 - 294}}{9.8} = \dfrac{20 \pm \sqrt{106}}{9.8}$, so $t = 0.990$ s or $t = 3.09$ s.

③The ball passes $15$ m on the way up and on the way down. Time above $15$ m $= \dfrac{2\sqrt{106}}{9.8} = 2.10$ s (3 s.f.).

5 Multi-Stage Problems

If the acceleration changes part-way through, split the motion into stages. Each stage has its own constant acceleration; the final velocity of one stage is the initial velocity of the next.

Worked Example 7 — Two stages

A car starts from rest and accelerates at $2$ m s$^{-2}$ for $10$ s. It then decelerates uniformly, coming to rest in a further $50$ m. Find the deceleration and the total time.

①Stage 1: $v = 0 + 2(10) = 20$ m s$^{-1}$; $s = \tfrac12(2)(10^2) = 100$ m.

②Stage 2: $u = 20$, $v = 0$, $s = 50$: $0 = 400 + 100a \Rightarrow a = -4$, a deceleration of $4$ m s$^{-2}$.

③$0 = 20 - 4t \Rightarrow t = 5$ s. Total time $= 10 + 5 = 15$ s.

Worked Example 8 — Two particles meeting

Particle $P$ is dropped from a height of $50$ m. At the same moment, particle $Q$ is projected vertically upwards from the ground directly below at $25$ m s$^{-1}$. Find when and at what height they meet.

①Height of $P$: $50 - 4.9t^2$. Height of $Q$: $25t - 4.9t^2$.

②Equal when $50 - 4.9t^2 = 25t - 4.9t^2 \Rightarrow 25t = 50 \Rightarrow t = 2$ s.

③Height $= 50 - 4.9(4) = 30.4$ m.

Both particles have the same acceleration, so the $4.9t^2$ terms cancel — relative to each other they move at a constant $25$ m s$^{-1}$, closing a $50$ m gap in $2$ s.

6 Quick Reference

No $s$

$v = u + at$

No $a$

$s = \tfrac12(u + v)t$

No $v$

$s = ut + \tfrac12 at^2$

No $u$

$s = vt - \tfrac12 at^2$

No $t$

$v^2 = u^2 + 2as$

Gravity

$a = -9.8$ with up positive, throughout the flight.

Greatest height

$v = 0$; height $= \dfrac{u^2}{2g}$.

Multi-stage

End velocity of one stage = start velocity of the next.

Accuracy

Using $g = 9.8$: give answers to 2 or 3 s.f.

7 Practice Questions

Take $g = 9.8$ m s$^{-2}$ and give non-exact answers to 3 significant figures.

Question 1

A particle has initial velocity $4$ m s$^{-1}$ and constant acceleration $3$ m s$^{-2}$. Find its velocity and displacement after $6$ s.

▶ Show solution

$v = 4 + 3(6) = 22$ m s$^{-1}$.

$s = 4(6) + \tfrac12(3)(36) = 24 + 54 = 78$ m.

Question 2

A train travelling at $25$ m s$^{-1}$ stops with uniform deceleration in $50$ m. Find the deceleration and the time taken.

▶ Show solution

$0 = 625 + 2a(50) \Rightarrow a = -6.25$: deceleration $6.25$ m s$^{-2}$.

$s = \tfrac12(u + v)t$: $50 = \tfrac12(25)t \Rightarrow t = 4$ s.

Question 3

A cyclist passes a post at $8$ m s$^{-1}$ and accelerates uniformly, covering $60$ m in the next $5$ s. Find the acceleration and the final speed.

▶ Show solution

$60 = 8(5) + \tfrac12 a(25) = 40 + 12.5a \Rightarrow a = 1.6$ m s$^{-2}$.

$v = 8 + 1.6(5) = 16$ m s$^{-1}$.

Question 4

A stone is dropped from rest from a bridge $45$ m above a river. Find the time it takes to reach the water and its speed on impact.

▶ Show solution

Down positive: $u = 0$, $a = 9.8$, $s = 45$.

$45 = 4.9t^2 \Rightarrow t = \sqrt{\dfrac{45}{4.9}} = 3.03$ s.

$v^2 = 0 + 2(9.8)(45) = 882 \Rightarrow v = 29.7$ m s$^{-1}$.

Question 5

A ball is thrown vertically upwards at $21$ m s$^{-1}$. Find the greatest height and the total time before it returns to the point of projection.

▶ Show solution

At the top $v = 0$: $0 = 441 - 19.6s \Rightarrow s = 22.5$ m.

Returns when $s = 0$: $0 = 21t - 4.9t^2 \Rightarrow t = \dfrac{21}{4.9} = 4.29$ s (exactly $\tfrac{30}{7}$ s).

Question 6

A particle moving with constant acceleration travels $18$ m in the first $2$ s and $30$ m in the next $2$ s. Find its initial velocity and acceleration.

▶ Show solution

First $2$ s: $18 = 2u + 2a$, so $u + a = 9$.

First $4$ s: $48 = 4u + 8a$, so $u + 2a = 12$.

Subtract: $a = 3$ m s$^{-2}$, then $u = 6$ m s$^{-1}$.

Question 7

A ball is thrown vertically downwards at $5$ m s$^{-1}$ from the top of a tower $30$ m high. Find the time to reach the ground and the speed on impact.

▶ Show solution

Down positive: $u = 5$, $a = 9.8$, $s = 30$.

$30 = 5t + 4.9t^2 \Rightarrow 4.9t^2 + 5t - 30 = 0$.

$t = \dfrac{-5 + \sqrt{25 + 588}}{9.8} = \dfrac{-5 + \sqrt{613}}{9.8} = 2.02$ s.

$v^2 = 25 + 2(9.8)(30) = 613 \Rightarrow v = 24.8$ m s$^{-1}$.

Question 8

A ball is projected vertically upwards at $15$ m s$^{-1}$. Find the length of time for which it is more than $10$ m above the point of projection.

▶ Show solution

$10 = 15t - 4.9t^2 \Rightarrow 4.9t^2 - 15t + 10 = 0$.

$t = \dfrac{15 \pm \sqrt{225 - 196}}{9.8} = \dfrac{15 \pm \sqrt{29}}{9.8}$, giving $t = 0.981$ s and $t = 2.08$ s.

Time above $10$ m $= \dfrac{2\sqrt{29}}{9.8} = 1.10$ s.

Question 9

Starting from the derivation $v = u + at$ and $s = \tfrac12(u + v)t$, show that $v^2 = u^2 + 2as$.

▶ Show solution

From $v = u + at$, $t = \dfrac{v - u}{a}$.

Substitute: $s = \tfrac12(u + v)\cdot\dfrac{v - u}{a} = \dfrac{v^2 - u^2}{2a}$.

So $2as = v^2 - u^2$, i.e. $v^2 = u^2 + 2as$. $\blacksquare$

Question 10

A model rocket is launched vertically from rest on the ground. Its engine gives it a constant upward acceleration of $4$ m s$^{-2}$ for $5$ s, after which the engine stops and the rocket moves freely under gravity.

(a) Find the speed and height of the rocket when the engine stops. (b) Find the greatest height reached. (c) Find the total time from launch until the rocket hits the ground. (d) State one modelling assumption you have made.

▶ Show solution

(a) $v = 0 + 4(5) = 20$ m s$^{-1}$; height $= \tfrac12(4)(25) = 50$ m.

(b) Free stage, up positive: $u = 20$, $a = -9.8$, $v = 0$.

$0 = 400 - 19.6s \Rightarrow s = 20.4$ m. Greatest height $= 50 + 20.4 = 70.4$ m.

(c) Free stage from the cut-out point to the ground: $s = -50$.

$-50 = 20t - 4.9t^2 \Rightarrow 4.9t^2 - 20t - 50 = 0$

$t = \dfrac{20 + \sqrt{400 + 980}}{9.8} = \dfrac{20 + \sqrt{1380}}{9.8} = 5.83$ s.

Total time $= 5 + 5.83 = 10.8$ s.

(d) For example: air resistance is ignored; the rocket is modelled as a particle; the loss of mass as fuel burns is ignored.

Constant Acceleration (suvat) · A Level Maths Year 1 · Created with MathJax