3.02MechanicsCore

Motion in a Straight Line

Q: How do I choose which suvat equation to use?

Identify which of s, u, v, a, t is neither given nor wanted, and use the equation without it.

Q: Can suvat give negative time?

A quadratic in t can have a negative root. It usually corresponds to a time before the motion started and is rejected — but say so.

When acceleration is constant — a car braking steadily, a stone falling freely — five equations connect displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ . Each equation leaves out one of the five quantities, so if you know three, one equation will always give you a fourth. These are the $\textbf{suvat}$ equations.

45 min Kinematics

The big picture

The suvat equations are what calculus gives when acceleration is constant: integrate $a$ once to get $v = u + at$ , again to get $s = ut + \tfrac12at^{2}$ , and eliminate $t$ or $u$ to get the rest. So they are not five facts to memorise but one idea written five ways. The real skill is modelling: choosing a positive direction, listing the known and wanted quantities, spotting which quantity is irrelevant, and recognising where one stage of motion ends and another begins. Vertical motion under gravity is the classic application, with $a = -g$ throughout the flight, and the same thinking extends directly to projectiles in two dimensions.

What you'll be able to do

Derive and use the five constant-acceleration equations
Select the appropriate equation by identifying known and unknown quantities
Solve problems involving vertical motion under gravity
Handle multi-stage motion and two objects moving in the same line

The suvat equations

For motion in a straight line with $\textbf{constant acceleration}$ : $s$ is displacement, $u$ initial velocity, $v$ final velocity, $a$ acceleration, $t$ time.

Integrating $a$ with $v = u$ at $t = 0$ gives $v = u + at$ ; integrating again with $s = 0$ at $t = 0$ gives $s = ut + \tfrac12at^{2}$ . The others follow by elimination.

List $s, u, v, a, t$ , fill in what is known, mark what is wanted, and choose the equation that does not contain the fifth quantity.

v = u + at

s

s = \tfrac12(u + v)t

a

s = ut + \tfrac12at^{2}

v

s = vt - \tfrac12at^{2}

u

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

t

Example 1A train decelerates uniformly from 30 m s⁻¹ to 12 m s⁻¹ over a distance of 420 m. Find the deceleration and the time taken.

u = 30

v = 12

s = 420

; want

a

and

t

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

144 = 900 + 840a

, so

a = -0.9

s = \tfrac12(u + v)t

420 = 21t

, so

t = 20

AnswerDeceleration

0.9

m s⁻²; time

20

Tip — A deceleration is a negative acceleration. Write $a = -0.9$ in the equations, but you may quote "deceleration 0.9 m s⁻²" in the answer.

Vertical motion under gravity

Objects moving freely under gravity have constant acceleration $g = 9.8$ m s⁻² downwards, if air resistance is ignored. OCR uses $g = 9.8$ unless told otherwise.

Taking upwards as positive gives $a = -9.8$ for the whole flight — on the way up, at the top, and on the way down.

At the highest point the velocity is momentarily zero, but the acceleration is still $-9.8$ .

Example 1A ball is thrown vertically upwards at 14.7 m s⁻¹ from a point 2 m above the ground. Find the greatest height above the ground and the time until it hits the ground.

1Up positive,

a = -9.8

. At the top

v = 0

0 = 14.7^{2} - 19.6s

, so

s = 11.025

m above the launch point.

2Greatest height above the ground:

13.025

3Ground is

s = -2

-2 = 14.7t - 4.9t^{2}

, so

4.9t^{2} - 14.7t - 2 = 0

t = \dfrac{14.7 + \sqrt{216.09 + 39.2}}{9.8} = \dfrac{14.7 + 15.978}{9.8} = 3.13

s (reject the negative root).

AnswerGreatest height

13.0

m; hits the ground after

3.13

Why this works

Solving for the whole flight in one equation works because acceleration is the same throughout. There is no need to split into "up" and "down" stages — just use the correct signed displacement for the end point.

Multi-stage motion

If acceleration changes part-way through, split the motion into stages with constant acceleration in each. The final velocity of one stage is the initial velocity of the next.

A velocity–time sketch is often the fastest way to organise a multi-stage problem, since distances are areas.

Example 1A cyclist accelerates from rest at 0.5 m s⁻² for 12 s, then continues at constant speed. How far has she travelled after 40 s?

1Stage 1:

v = 0 + 0.5 \times 12 = 6

m s⁻¹ and

s = \tfrac12 \times 0.5 \times 144 = 36

2Stage 2: 28 s at 6 m s⁻¹ gives

168

Answer

204

Two objects in the same line

When two objects move along the same line, write each one’s displacement from a common origin as a function of time, then set them equal to find when they meet.

Take care if they start at different times — the later one has been moving for $t - t_{0}$ seconds.

Example 1Car A passes a point

P

at a constant 20 m s⁻¹. At that instant car B starts from rest at

P

and accelerates at 2.5 m s⁻² in the same direction. When and where does B catch A?

1A:

s = 20t

. B:

s = 1.25t^{2}

1.25t^{2} = 20t

gives

t = 0

t = 16

3At

t = 16

s = 320

AnswerAfter 16 s, 320 m from

P

Tip — The solution $t = 0$ is the moment they are together at the start. Always interpret each root rather than discarding one silently.

Think like an examiner

Write out $s, u, v, a, t$ with known values before choosing an equation.
State the positive direction and sign $g$ accordingly.
Use $g = 9.8$ m s⁻² and give answers to 3 significant figures unless told otherwise.
Comment on modelling assumptions such as air resistance where relevant.

Common misconceptions

Using suvat when acceleration varies with time.

Suvat only applies to constant acceleration; otherwise use calculus.

Taking

a = +9.8

for a ball thrown upwards with up as positive.

Gravity acts downwards, so

a = -9.8

Saying acceleration is zero at the top of the flight.

Velocity is zero there; acceleration is still

-g

Using

s = 0

for the ground when launching from a height.

Measure displacement from the launch point: the ground is at

s = -h

Constant acceleration

v = u + at

s

s = ut + \tfrac12at^{2}

v

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

t

s = \tfrac12(u + v)t

a

Stretch yourself

A stone is dropped from rest at the top of a cliff. One second later a second stone is thrown vertically downwards from the same point at 12 m s⁻¹. Both reach the sea at the same instant. Find the height of the cliff. Take $g = 9.8$ m s⁻².

Hint — Let the first stone fall for $t$ seconds; the second falls for $t - 1$ seconds. Equate their displacements.

Reveal solution

SolutionTaking down as positive: first stone

s = 4.9t^{2}

; second stone

s = 12(t - 1) + 4.9(t - 1)^{2}

. Equating:

4.9t^{2} = 12t - 12 + 4.9t^{2} - 9.8t + 4.9

, so

0 = 2.2t - 7.1

and

t = \dfrac{7.1}{2.2} = 3.227

s. Height

= 4.9 \times 3.227^{2} = 51.0

m. Check with the second stone: it falls for

2.227

s, covering

12 \times 2.227 + 4.9 \times 2.227^{2} = 26.73 + 24.31 = 51.0

m ✓.

\blacksquare

Questions students ask

How do I choose which suvat equation to use?

Identify which of

s, u, v, a, t

is neither given nor wanted, and use the equation without it.

Why is

g

sometimes 9.8 and sometimes 9.81?

OCR uses 9.8 m s⁻² unless a question states otherwise. Always use the value given.

Can suvat give negative time?

A quadratic in

t

can have a negative root. It usually corresponds to a time before the motion started and is rejected — but say so.

What assumptions are made in free-fall problems?

The object is a particle, air resistance is negligible, and

g

is constant over the height involved.

Key takeaways

Five suvat equations apply only to constant acceleration.
List known and wanted quantities; use the equation without the fifth.
Under gravity, $a = -9.8$ (up positive) throughout the motion.
Split multi-stage motion; velocity carries over between stages.
For two objects, equate displacements from a common origin.

How this fits the course

Build on

Displacement, Velocity and Acceleration

Newton’s Laws of Motion

Leads to

Projectile Motion

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

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Put the methods into practice: see how A-Level maths duels work. An account is required to play; the overview and sample are public.