Equations for uniformly accelerated motion

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Everything in mechanics starts with describing motion: where a body is, how fast it is going and how quickly that is changing.

This section opens the motion part of Unit 1 (WPH11, Mechanics and Materials), the 1 hour 30 minute written paper.

The same equations come back in the practical papers: Unit 3 (WPH13) can ask you to process light-gate or free-fall data with them.

You need to be able to use the equations for uniformly accelerated motion in one dimension: $s = \dfrac{(u + v)t}{2}$ , $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ .

How the marks work

Each marking point is one mark, written (1). A calculation earns a "use of" mark for each equation you write with your numbers substituted, then one mark for the final value with its unit.

A missing or wrong unit on a final answer usually loses the last mark (once per question part; not on a "show that", and not where the unit is printed). Using $g = 10$ m s⁻² also loses a mark: use $9.81$ m s⁻².

An earlier wrong value carries forward (ecf), so later marks are still open on your own number. Some marks are dependent: a conclusion scores only after a valid calculation.

A "show that" answer must go at least one significant figure beyond the printed value. Working backwards from the printed value can earn full marks on a 2-mark item, but at most 2 marks on a 3-mark item.

"Deduce" items end with a separate comparison of two values and a conclusion in words; that last mark is the one most often lost.

The one asterisked 6-mark question per paper scores up to 4 marks for indicative points and up to 2 for linking them into a sustained line of reasoning.

Learn it: not on the list

The list of data, formulae and relationships at the back of the paper prints all four equations above and $g = 9.81$ m s⁻².

What it does not print, and you must know: average velocity $= \dfrac{\Delta s}{\Delta t}$ and acceleration $= \dfrac{\Delta v}{\Delta t}$ as definitions; what the gradients and areas of motion graphs give (later in this unit); resolving a vector with $\sin$ and $\cos$ ; Pythagoras and $\tan$ for a resultant at right angles; and the rule that the horizontal and vertical motions of a projectile are independent.

The idea: five quantities describe straight-line motion

For a body moving along a straight line with a constant acceleration, five quantities tell the whole story of one stage of its motion:

Symbol

Quantity

Unit

$s$

displacement

$u$

initial velocity

m s⁻¹

$v$

final velocity

m s⁻¹

$a$

acceleration

m s⁻²

$t$

time taken

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

An acceleration of 2.0 m s⁻² means the velocity grows by 2.0 m s⁻¹ every second.

"Uniform" means the acceleration stays the same for the whole stage.

Each equation links four of the five quantities, so if you know any three you can find the other two.

Displacement, velocity and acceleration all have a direction (they are vectors, as the section on scalars and vectors explains).

On a straight line the direction is just a sign: choose one way as positive, and anything pointing the other way is negative.

A body that is slowing down has an acceleration opposite to its velocity, so if its velocity is positive its acceleration is negative.

Why the equations are true

Picture the motion on a velocity-time graph.

Uniform acceleration from $u$ to $v$ in time $t$ is a straight line.

Velocity against time for a body accelerating uniformly from u to v in time t. The area under the straight line is split into a rectangle of area ut and a triangle of area one half (v minus u) t.

1. The gradient of the line is the acceleration: $a = \dfrac{v - u}{t}$ . Rearranged, $v = u + at$ .

2. The area under the line is the displacement (the section on gradients and areas explains why). The area is a trapezium, so $s = \dfrac{(u + v)}{2} \times t$ . In words: with uniform acceleration, the average velocity is halfway between $u$ and $v$ .

3. Split the same area into a rectangle and a triangle: $s = ut + \tfrac{1}{2}(v - u)t$ . Since $v - u = at$ , this gives $s = ut + \tfrac{1}{2}at^2$ .

4. Remove $t$ : from step 1, $t = \dfrac{v - u}{a}$ . Put that into step 2: $s = \dfrac{(u + v)(v - u)}{2a} = \dfrac{v^2 - u^2}{2a}$ , so $v^2 = u^2 + 2as$ .

You will not be asked to derive these on WPH11 (they are printed), but knowing where they come from tells you when they fail.

Every step used a straight velocity-time line.

If the acceleration changes (a car whose engine pulls less hard as it speeds up, a skydiver with growing drag), the line curves and none of the four equations is exact.

You then go back to the graph itself, reading gradients and areas.

Choosing the right equation

Each equation leaves out one of the five quantities.

Write down the three you know and the one you want; the equation to use is the one that leaves out the fifth.

Equation

Quantity it leaves out

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

$a$

$v = u + at$

$s$

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

$v$

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

$t$

Before substituting, put every value in base SI units.

Papers deliberately mix prefixes and non-SI units: cm, mm, km, ms, minutes and km h⁻¹ all appear.

To convert km h⁻¹ to m s⁻¹, divide by 3.6 (1 km h⁻¹ $= \dfrac{1000\ \text{m}}{3600\ \text{s}}$ ).

What the equations tell you, not only the numbers

The equation $v^2 = u^2 + 2as$ with $v = 0$ gives a braking distance $s = \dfrac{u^2}{2a}$ for a deceleration of size $a$ .

The braking distance depends on the square of the speed: double the speed and, with the same brakes, the car needs four times the distance to stop.

That single line is the physics behind lower speed limits near schools.

In the same way, $s = \tfrac{1}{2}at^2$ from rest says the distance covered grows with the square of the time: in the second second of a drop a stone falls three times as far as in the first.

Key terms

Displacement: the distance moved in a straight line from the starting point to the finishing point, in a stated direction.

Velocity: the rate of change of displacement.

Acceleration: the rate of change of velocity.

Uniform acceleration: an acceleration that stays constant, so the velocity changes by equal amounts in equal times.

Worked example 1: a direct substitution

A train speeds up uniformly from 12 m s⁻¹ to 30 m s⁻¹ in 45 s.

Calculate its acceleration.

1. Known: $u = 12$ m s⁻¹, $v = 30$ m s⁻¹, $t = 45$ s; want $a$ . Use $v = u + at$ , which leaves out $s$ .

2. $30 = 12 + a \times 45$ (1) use of v = u + at

3. $a = \dfrac{30 - 12}{45} = 0.40$ m s⁻² (1) value with unit

Answer: 0.40 m s⁻².

Its velocity rises by 0.40 m s⁻¹ every second, which is gentle enough for standing passengers.

Worked example 2: choosing the equation that leaves out time

An aircraft starts from rest and accelerates uniformly along a runway at 2.6 m s⁻².

It leaves the ground at 72 m s⁻¹.

Calculate the length of runway it uses.

1. Known: $u = 0$ , $v = 72$ m s⁻¹, $a = 2.6$ m s⁻²; want $s$ . Time is not given or asked for, so use $v^2 = u^2 + 2as$ .

2. $72^2 = 0 + 2 \times 2.6 \times s$ (1) use of v² = u² + 2as

3. $s = \dfrac{5184}{5.2} = 997$ m (1) value with unit

Answer: 997 m, so just under 1 km of runway, before any safety margin.

Worked example 3: a show that, then a follow-on part

A cyclist stops pedalling and slows down uniformly.

She travels 35 m in 6.0 s, and her speed at the start is 8.4 m s⁻¹.

(a) Show that her speed at the end of the 6.0 s is about 3 m s⁻¹. (2)

1. $s = \dfrac{(u + v)t}{2}$ : $35 = \dfrac{(8.4 + v) \times 6.0}{2}$ (1) use of s = (u + v)t/2

2. $v = \dfrac{2 \times 35}{6.0} - 8.4 = 3.27$ m s⁻¹ (1) value to at least one more significant figure than "about 3"

(b) Calculate her deceleration. (2)

3. $a = \dfrac{v - u}{t} = \dfrac{3.27 - 8.4}{6.0}$ (1) use of v = u + at (ecf from (a))

4. $a = -0.856$ m s⁻², so the deceleration is 0.86 m s⁻² (1) value with unit

Answer: (a) 3.27 m s⁻¹; (b) 0.86 m s⁻².

Writing only "3 m s⁻¹" in (a) would lose the second mark.

In (b), keep the unrounded 3.267 in your calculator: rounding it to 3 first would give 0.90 m s⁻².

Worked example 4: read the prefix

A trolley accelerates uniformly down a ramp.

Its velocity rises from 15 cm s⁻¹ to 85 cm s⁻¹ while it moves 60 cm.

Calculate its acceleration.

1. Convert first: $u = 0.15$ m s⁻¹, $v = 0.85$ m s⁻¹, $s = 0.60$ m. No time is given, so use $v^2 = u^2 + 2as$ .

2. $0.85^2 = 0.15^2 + 2 \times a \times 0.60$ (1) use of v² = u² + 2as

3. $a = \dfrac{0.7225 - 0.0225}{1.20} = 0.58$ m s⁻² (1) value with unit

Answer: 0.58 m s⁻².

Left in centimetres the arithmetic gives 58, which is the right number in cm s⁻², but the paper expects SI units.

Connect it

These four equations carry most of the motion questions in this unit: free fall and thrown objects next, then each direction of a projectile's flight.

In unit 1.3b they finish almost every force question: $\Sigma F = ma$ gives the acceleration, and these equations turn it into a speed or a distance.

Unit 4 uses them again for charged particles accelerated in a uniform electric field.

Road-safety engineers use the same $v^2 = u^2 + 2as$ to set stopping distances.

Exam tip

Before you calculate, write the five letters $s, u, v, a, t$ in a column and fill in what you know, with signs and SI units.

Then write the equation you chose in symbols, the same equation with numbers substituted, and the answer with its unit, each on its own line.

The substituted line is what earns the "use of" mark, even if a slip follows later.

Common mistakes

Using the equations when the acceleration is not constant. They come from a straight velocity-time line. For a curved line (changing acceleration), use the graph's gradient and area instead.

Using speed = distance ÷ time for a body that is speeding up. That gives the average velocity, not the final velocity. With uniform acceleration from rest the final velocity is twice the average.

Leaving units mixed. Convert km h⁻¹, cm, mm and ms to m s⁻¹, m and s before substituting.

Losing the sign of a deceleration. If the velocity is positive and the body is slowing, $a$ is negative in the equation. Quote a "deceleration" as a positive number in the answer.

Check your understanding

1. A car travelling at speed $u$ brakes to rest with a uniform deceleration of size $a$ . Which of the following gives its braking distance?

A $\dfrac{u^2}{2a}$

B $\dfrac{u}{2a}$

C $\dfrac{2a}{u^2}$

D $\dfrac{u^2}{a}$

(1) Total [1]

2. A lift starts from rest and accelerates uniformly to 2.4 m s⁻¹ in 1.6 s.

(a) Calculate the acceleration of the lift. (2)

(b) Calculate the distance the lift travels in this time. (2) Total [4]

3. An electric scooter is moving at 3.0 m s⁻¹. It then accelerates uniformly at 0.80 m s⁻² over a distance of 12 m. Show that its speed at the end of this distance is about 5 m s⁻¹. (2) Total [2]

4. A sprinter starts from rest and accelerates uniformly, reaching 9.6 m s⁻¹ after running 20 m. Calculate the time taken and her acceleration. (3) Total [3]

5. A driver travelling at 22 m s⁻¹ sees a fallen tree 48 m ahead. Her reaction time is 0.70 s, and the car then decelerates uniformly at 6.5 m s⁻². Deduce whether the car stops before it reaches the tree. (4) Total [4]

6. A student says: "The equation $s = \dfrac{(u + v)t}{2}$ works for any motion, because it is just average speed multiplied by time." Explain why the student is wrong. (2) Total [2]

Answers

1. A (1). From $v^2 = u^2 + 2as$ with $v = 0$ and acceleration $-a$ : $s = \dfrac{u^2}{2a}$ . B has $u$ not squared. C is the reciprocal. D misses the factor of 2.

2. (a) $a = \dfrac{v - u}{t} = \dfrac{2.4 - 0}{1.6}$ (1); $a = 1.5$ m s⁻² (1).

(b) $s = \dfrac{(u + v)t}{2} = \dfrac{(0 + 2.4) \times 1.6}{2}$ (1); $s = 1.92$ m (1). Allow $s = \tfrac{1}{2}at^2$ with ecf from (a).

3. $v^2 = u^2 + 2as = 3.0^2 + 2 \times 0.80 \times 12$ (1); $v = \sqrt{28.2} = 5.31$ m s⁻¹ (1). The answer must go beyond the printed 5.

4. $s = \dfrac{(u + v)t}{2}$ gives $20 = \dfrac{9.6\,t}{2}$ (1); $t = 4.2$ s (4.17 s) (1); $a = \dfrac{9.6}{4.17} = 2.3$ m s⁻² (1). Allow $v^2 = 2as$ for $a$ .

5. Distance during the reaction time $= 22 \times 0.70 = 15.4$ m (1); braking distance from $v^2 = u^2 + 2as$ : $0 = 22^2 - 2 \times 6.5 \times s$ (1), so $s = 37.2$ m (1); total stopping distance $= 52.6$ m, which is more than 48 m, so the car does not stop before the tree (1). The last mark needs both values compared and the conclusion.

6. The equation uses the average velocity $\dfrac{u + v}{2}$ , which is halfway between $u$ and $v$ only when the acceleration is uniform (a straight velocity-time line) (1). If the acceleration changes, the average velocity is not halfway between the start and end values, so the equation gives the wrong displacement (1).

Equations for uniformly accelerated motion

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