Q3MechanicsCore

Constant Acceleration

Q: How do I pick the right suvat equation?

List s, u, v, a, t, mark what is known and wanted, and find the one quantity that is neither. The equation omitting it is the one to use.

When acceleration is constant, the velocity–time graph is a straight line — and everything about the motion can be read off it. The four $\textbf{suvat}$ equations are that geometry written algebraically, and choosing between them is a matter of noticing which quantity you do not need.

35 min Kinematics

The big picture

These four equations solve an enormous share of mechanics problems, and the temptation is to memorise them and hunt for the one that fits. Deriving them once from the graph is a better investment: each equation is the trapezium area or the gradient rearranged, and seeing that makes the set feel like one idea rather than four facts. It also makes the crucial restriction memorable — the derivation uses a straight line, so the moment acceleration varies the equations are simply wrong, and you need the calculus of the fourth lesson instead.

What you'll be able to do

Derive the suvat equations from a velocity–time graph
Select the appropriate equation by identifying the unused quantity
Solve problems including those requiring a quadratic
Handle multi-stage journeys by splitting into constant-acceleration phases

Where the equations come from

For constant acceleration the velocity–time graph is a straight line from $u$ at time 0 to $v$ at time $t$ . Its gradient is the acceleration, which gives $a = \dfrac{v-u}{t}$ and hence $v = u + at$ .

The area under that line is the displacement. As a trapezium with parallel sides $u$ and $v$ and width $t$ , the area is $\dfrac{u+v}{2}\,t$ — the third equation, and the one that says displacement equals average velocity times time.

Splitting the trapezium into a rectangle of height $u$ and a triangle of height $at$ gives $s = ut + \tfrac12 at^{2}$ . Eliminating $t$ between the first and third gives $v^{2} = u^{2} + 2as$ .

All four therefore encode the same straight-line graph. They differ only in which of the five quantities has been eliminated.

v = u + at

From the gradient. Omits

s

s = \left(\dfrac{u+v}{2}\right)t

Trapezium area. Omits

a

s = ut + \tfrac{1}{2}at^{2}

Rectangle plus triangle. Omits

v

v^{2} = u^{2} + 2as

Eliminating

t

. Omits

t

Why this works

The third equation, $s = \left(\tfrac{u+v}{2}\right)t$ , only works because the graph is straight — the average of the endpoints equals the mean velocity for a linear graph and not otherwise. That single observation is why the whole set requires constant acceleration.

Choosing the equation

List the five quantities $s$ , $u$ , $v$ , $a$ , $t$ and mark which are known and which is wanted. Exactly one will be neither — and the equation that omits it is the one to use.

Writing the list out takes a few seconds and removes all guesswork. It also makes the working legible, which matters when a question has several parts building on each other.

Example 1A car accelerates uniformly from

8\ \mathrm{m\,s^{-1}}

20\ \mathrm{m\,s^{-1}}

over a distance of 84 m. Find the acceleration.

1Known:

u = 8

v = 20

s = 84

. Wanted:

a

. Not needed:

t

2The equation without

t

v^{2} = u^{2} + 2as

400 = 64 + 168a

, so

336 = 168a

Answer

a = 2\ \mathrm{m\,s^{-2}}

Example 2A particle starting from rest accelerates at

3\ \mathrm{m\,s^{-2}}

. How long to travel 96 m?

1Known:

u = 0

a = 3

s = 96

. Wanted:

t

. Not needed:

v

2Use

s = ut + \tfrac12at^{2}

96 = 0 + 1.5t^{2}

t^{2} = 64

Answer

t = 8\ \mathrm{s}

Tip — Write out $s, u, v, a, t$ every time. The gap in the list picks the equation, so there is never a need to try several.

When a quadratic appears

Using $s = ut + \tfrac12 at^{2}$ to find $t$ produces a quadratic, which may have two positive roots. Both can be physically meaningful.

A projectile passing a given height on the way up and again on the way down gives two valid times. A negative root, however, refers to a time before the motion began and is discarded.

When two positive roots appear, read the question to decide whether it wants the first, the second, or both.

Example 1A particle has

u = 12\ \mathrm{m\,s^{-1}}

and

a = -4\ \mathrm{m\,s^{-2}}

. Find the times at which its displacement is 8 m.

8 = 12t - 2t^{2}

, so

2t^{2} - 12t + 8 = 0

, i.e.

t^{2} - 6t + 4 = 0

2Quadratic formula:

t = \dfrac{6 \pm \sqrt{36-16}}{2} = \dfrac{6 \pm \sqrt{20}}{2}

\sqrt{20} = 4.472

, giving

t = 0.764

t = 5.236

Answer

t = 0.76

s and

t = 5.24

s — passing 8 m on the way out and again on the way back.

Why this works

The two roots are not an ambiguity to be resolved but a genuine feature of the motion: with negative acceleration the particle reaches 8 m, continues, turns around, and returns through 8 m. Discarding one without reason loses a mark.

Multi-stage journeys

The equations require acceleration to be constant, so a journey with several phases must be split at each change. Solve each phase separately.

The link between phases is that the final velocity of one is the initial velocity of the next, and displacements add. Setting the working out phase by phase, with a clear label for each, keeps it manageable.

A velocity–time sketch is often the fastest way to see the structure before any algebra.

Example 1A train accelerates from rest at

0.5\ \mathrm{m\,s^{-2}}

for 40 s, travels at constant speed for 100 s, then decelerates uniformly to rest in 25 s. Find the total distance.

1Phase 1:

v = 0 + 0.5(40) = 20\ \mathrm{m\,s^{-1}}

;

s = \tfrac12(0.5)(1600) = 400

2Phase 2: constant

20\ \mathrm{m\,s^{-1}}

for 100 s gives

2000

3Phase 3: from 20 to 0 in 25 s, so

s = \left(\tfrac{20+0}{2}\right)(25) = 250

4Total:

400 + 2000 + 250

Answer

2650\ \mathrm{m}

Tip — Never apply a single suvat equation across a change in acceleration. Splitting into phases is not optional — the equations are simply invalid otherwise.

Think like an examiner

List $s, u, v, a, t$ and state which equation you are using and why.
Take care with signs: a deceleration is a negative acceleration relative to the direction of motion.
When a quadratic gives two positive roots, interpret both rather than silently discarding one.
Split multi-stage journeys explicitly and label each phase.

Common misconceptions

Applying suvat across a change in acceleration.

The equations assume constant acceleration. Split the journey into phases.

Discarding a positive root without checking.

Two positive times often both occur. Only negative times are automatically rejected.

Using

s = \left(\tfrac{u+v}{2}\right)t

for varying acceleration.

That average only equals the mean velocity for a straight-line graph.

Forgetting that "starts from rest" means

u = 0

It is given information. Similarly "comes to rest" means

v = 0

The suvat equations

v = u + at

Omits

s

s = ut + \tfrac12 at^{2}

Omits

v

s = \left(\tfrac{u+v}{2}\right)t

Omits

a

v^{2} = u^{2}+2as

Omits

t

\text{constant } a \text{ only}

Otherwise use calculus.

Stretch yourself

A car travelling at $30\ \mathrm{m\,s^{-1}}$ passes a stationary police motorcycle, which immediately gives chase, accelerating uniformly at $4\ \mathrm{m\,s^{-2}}$ . Find how long the motorcycle takes to catch the car, and how fast it is then going.

Hint — They meet when their displacements are equal. Write an expression for each and set them equal.

Reveal solution

SolutionMeasure both displacements from the moment of passing. The car has constant velocity, so

s_{\text{car}} = 30t

. The motorcycle starts from rest with

a = 4

, so

s_{\text{bike}} = \tfrac12(4)t^{2} = 2t^{2}

. They meet when

2t^{2} = 30t

, i.e.

2t(t - 15) = 0

, giving

t = 0

t = 15

s. The root

t = 0

is the initial moment when they were level, so the catch happens at

t = 15

s. The motorcycle’s speed then is

v = 0 + 4(15) = 60\ \mathrm{m\,s^{-1}}

— twice the car’s speed, which makes sense: to cover the same distance from rest in the same time under constant acceleration, its average speed must equal 30, and for uniform acceleration from rest the final speed is twice the average.

\blacksquare

Questions students ask

How do I pick the right suvat equation?

List

s

u

v

a

t

, mark what is known and wanted, and find the one quantity that is neither. The equation omitting it is the one to use.

What if the acceleration is not constant?

The suvat equations do not apply. Either split the motion into constant-acceleration phases, or use calculus as in the variable-acceleration lesson.

Why do some problems give two answers for time?

Because the object can be at the same displacement twice — once travelling outwards and once returning. Both positive roots are then physically real.

Is deceleration just negative acceleration?

It is acceleration opposing the current motion. If the velocity is positive then deceleration is negative acceleration, but if the velocity is negative a negative acceleration speeds the object up.

Key takeaways

Constant acceleration gives a straight velocity–time graph; the four equations encode it.
Each equation omits one of $s$ , $u$ , $v$ , $a$ , $t$ — the one you neither know nor want.
Quadratics can give two valid positive times; interpret rather than discard.
Split multi-stage journeys wherever the acceleration changes.
The equations are invalid the moment acceleration varies.