4.3.3Forces and Motion

SUVAT Equations

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4.3.3 SUVAT Equations — AQA A Level Physics 7408 notes: Select and apply the equations of motion for constant acceleration to solve problems in a straight line, including free fall and stopping distances.The five quantities in uniformly accelerated motion are:s displacement (m)u …

only when acceleration is constant
v = u + at
s = ½(u + v)t
s = ut + ½at2
v2 = u2 + 2as
thinking distance

Section 1 of 9Variables and conditions

The five quantities in uniformly accelerated motion are:

s displacement (m)

u initial velocity (m s⁻¹)

v final velocity (m s⁻¹)

a constant acceleration (m s⁻²)

t time (s)

The equations apply only when acceleration is constant and along a straight line. They cannot be used for motion with changing acceleration, such as a falling object with drag near terminal velocity.

Section 2 of 9The equations

v u + at
(no s).

From the definition of acceleration.
What the symbols mean: v = final velocity (m s⁻¹) · u = initial velocity (m s⁻¹) · a = acceleration (m s⁻²) · t = time (s)

½(u + v)t
(no a).

Average velocity multiplied by time, the area of a trapezium under the v–t graph.
What the symbols mean: s = displacement (m) · u = initial velocity · v = final velocity (m s⁻¹) · t = time (s); (u + v)/2 is the average velocity

s ut + ½at2
(no v).

Combine the first two.
What the symbols mean: s = displacement (m) · u = initial velocity (m s⁻¹) · a = acceleration (m s⁻²) · t = time (s)

v2 u2 + 2as
(no t).

Eliminate t.
What the symbols mean: v = final velocity · u = initial velocity (m s⁻¹) · a = acceleration (m s⁻²) · s = displacement (m)

(Also s = vt − ½at², which lacks u.) Each equation links four of the five variables, so you need three known quantities to find the others. Choose the equation which contains the unknown and the three knowns, and does not contain the quantity that is neither given nor wanted.

Section 3 of 9Method and sign convention

Write down s, u, v, a and t, marking the three you know and the one you need.

Choose a positive direction and keep to it.

A vector in the opposite direction is negative.
For a ball thrown up, with up positive, a = −9.8 m s⁻²; with down positive, a = +9.8 m s⁻².

Pick the equation that omits the unknown you don't need, substitute with units, and check the answer is sensible.

Where the motion has several stages (accelerate,

then cruise,
then brake), treat each separately; the final velocity of one stage becomes the initial velocity of the next.

Section 4 of 9Free fall and projectiles

Close to the Earth's surface, with air resistance negligible, all objects in free fall have the same downward acceleration g = 9.8 m s⁻².

For a body dropped from rest, u = 0, so s = ½gt² and v = gt.

For a ball thrown upwards, v = 0 at the top,

and the time up equals the time down when it returns to the same level, with the speed on return equal to the launch speed.

A projectile's horizontal and vertical motions are independent: horizontally, constant velocity (a = 0); vertically, constant acceleration g.

Use SUVAT separately in each direction, linked by the common time of flight.

Section 5 of 9Stopping distances

The stopping distance of a vehicle is the sum of the thinking distance (reaction time × speed, travelled at constant speed) and the braking distance (found from v² = u² + 2as with v = 0).

1Square of the initial speed

Since braking distance = u²/(2|a|), it is proportional to the square of the initial speed: doubling speed quadruples braking distance.

Wet roads, worn tyres and a heavier load reduce the deceleration and increase braking distance,

while tiredness or alcohol increases reaction time and thinking distance.

Section 6 of 9Choosing the right equation: worked strategy

Suppose a question says

a cyclist travelling at 4.0 m s⁻¹ accelerates uniformly and covers 60 m in 5.0 s;
- what is her final speed?

The knowns are u, s and t, and the unknown is v.

The equation with these four variables (and no a) is s = ½(u + v)t.

So 60 = ½(4.0 + v) × 5.0, which gives 4.0 + v = 24, so v = 20 m s⁻¹.

Often the same answer can be found in two ways, and doing one as a check is good practice.

Choosing the right equation: worked strategy

Where a question gives only two quantities, look for a hidden third. 'Starts from rest' means u = 0. 'Comes to rest' means v = 0. 'Dropped' means u = 0 and a = g. 'Thrown up and returns to the same height' means s = 0. 'At the top of its flight' means vertical v = 0.

For two-stage journeys, draw a velocity–time graph first. The area under it gives the total displacement without any algebra.

From the definition of acceleration, a = (v − u)/t gives v = u + at.

The displacement is the area under the v–t graph, a trapezium with parallel sides u and v and width t, so s = ½(u + v)t.

Substituting v = u + at gives s = ut + ½at².

Eliminating t by writing t = (v − u)/a in s = ½(u + v)t gives s = (v² − u²)/(2a), that is v² = u² + 2as.

Being able to derive these shows you understand where each equation comes from and the condition (constant acceleration) under which it holds.

Section 7 of 9Worked examples

A car accelerates uniformly from 12 m s⁻¹ to 28 m s⁻¹ over a distance of 200 m. Find the acceleration and the time taken.
A stone is thrown vertically upwards at 15 m s⁻¹ from a cliff top. It lands in the sea 4.0 s later. Find the height of the cliff. (g = 9.8 m s⁻².)
A car travelling at 25 m s⁻¹ brakes with constant deceleration 6.0 m s⁻² after a reaction time of 0.70 s. Calculate the total stopping distance.

Section 8 of 9Key terms

Term	Meaning
Uniform acceleration	Acceleration of constant magnitude and direction.
Free fall	Motion with only gravity acting.
Thinking distance	Distance travelled during the driver's reaction time.
Braking distance	Distance travelled while the brakes decelerate the vehicle.
Projectile	Object moving under gravity alone after launch.
g	Acceleration of free fall, 9.8 m s⁻².

Section 9 of 9Exam technique

Using SUVAT when acceleration is not constant.

Mixing the signs, for example using a = +9.8 for an upward throw with up as positive.

Substituting without converting units (km h⁻¹ to m s⁻¹).

Assuming the speed at the top of a throw is zero in both components of a projectile; only the vertical component is zero.

Forgetting the thinking distance when asked for the stopping distance.

Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at | s = ut + ½at² | v² = u² + 2as | s = ½(u + v)t

Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
Free Fall in vacuum: a = g ≈ 9.81 m/s² downwards (air resistance negligible).

Section 1 of 9Variables and conditions

Section 2 of 9The equations

Section 3 of 9Method and sign convention

Section 4 of 9Free fall and projectiles

Section 5 of 9Stopping distances

Section 6 of 9Choosing the right equation: worked strategy

Section 7 of 9Worked examples

A car accelerates uniformly from 12 m s−1 to 28 m s−1 over a distance of 200 m. Find the acceleration and the time taken.

A stone is thrown vertically upwards at 15 m s−1 from a cliff top. It lands in the sea 4.0 s later. Find the height of the cliff. (g = 9.8 m s−2.)

A car travelling at 25 m s−1 brakes with constant deceleration 6.0 m s−2 after a reaction time of 0.70 s. Calculate the total stopping distance.

Section 8 of 9Key terms

Section 9 of 9Exam technique

Marked practiceQuizzes for this lesson

Constant Acceleration Equations (SUVAT) & Slopes

A car accelerates uniformly from 12 m s⁻¹ to 28 m s⁻¹ over a distance of 200 m. Find the acceleration and the time taken.

A stone is thrown vertically upwards at 15 m s⁻¹ from a cliff top. It lands in the sea 4.0 s later. Find the height of the cliff. (g = 9.8 m s⁻².)

A car travelling at 25 m s⁻¹ brakes with constant deceleration 6.0 m s⁻² after a reaction time of 0.70 s. Calculate the total stopping distance.