A stone is dropped from rest and falls for 2.0 s. How far does it fall? [2 marks]

EnglandPhysicsSyllabus dot point

How do displacement, velocity and acceleration relate when motion is in a straight line?

Definitions of displacement, speed, velocity and acceleration, interpreting motion graphs, the equations of uniformly accelerated motion (suvat), and motion under gravity.

A focused answer to AQA A-Level Physics 3.4.1.3, covering displacement, speed, velocity and acceleration, the interpretation of displacement-time and velocity-time graphs, the suvat equations of uniformly accelerated motion, and motion under gravity.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Defining the quantities
Motion graphs
The suvat equations
Motion under gravity
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What this dot point is asking

AQA specification point 3.4.1.3 wants you to define the kinematic quantities, read and interpret displacement-time and velocity-time graphs, select and use the equations of uniformly accelerated motion, and apply them to objects moving freely under gravity.

Defining the quantities

Speed and distance are the scalar partners of velocity and displacement: they have magnitude but no direction. A car that drives a lap of a circuit covers a large distance but returns to a zero displacement, and its average speed is non-zero while its average velocity is zero.

Motion graphs

On a displacement-time graph the gradient gives the velocity; a straight line means constant velocity, and a curved line means changing velocity (acceleration). On a velocity-time graph the gradient gives the acceleration and the area under the line gives the displacement.

The instantaneous velocity at a moment is found from the gradient of the tangent to a displacement-time curve at that point, and the instantaneous acceleration from the gradient of the tangent to a velocity-time curve. The shapes are revealing: a horizontal line on a velocity-time graph means constant velocity (zero acceleration), a straight slope means constant acceleration, and a curve means changing acceleration. A negative gradient on a velocity-time graph means deceleration, and the area below the time axis counts as negative displacement, so an object that goes out and comes back has a net displacement found by subtracting the areas.

The suvat equations

For uniform (constant) acceleration, the four equations link displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ .

Pick the equation that contains the three quantities you know plus the one you want, so only one unknown remains. Each equation omits one of the five variables, so identifying which variable is not involved guides the choice.

Motion under gravity

Near the Earth's surface, an object in free fall has a constant downward acceleration $g \approx 9.81 \text{ m s}^{-2}$ (ignoring air resistance). The suvat equations apply directly with $a = g$ . An object thrown upwards decelerates, stops momentarily at the top, then accelerates back down, so a consistent sign convention (taking one direction as positive) is essential. A common method to measure $g$ uses a timed free fall over a measured drop with $s = \tfrac{1}{2}gt^2$ .

Finding the speed of a dropped stone using suvat

A stone is dropped from rest down a well and takes $1.8 \text{ s}$ to reach the water. Find its speed on impact and the depth of the well. Take $g = 9.81 \text{ m s}^{-2}$ .

step 1: List the known quantities

$u = 0$ , $a = g = 9.81 \text{ m s}^{-2}$ , $t = 1.8 \text{ s}$ , taking downwards as positive.

step 2: Find the impact speed

Use $v = u + at = 0 + (9.81)(1.8) = 17.7 \text{ m s}^{-1}$ .

step 3: Choose an equation for the depth

Use $s = ut + \tfrac{1}{2}at^2$ , with $u = 0$ .

step 4: Evaluate the depth

$s = 0 + \tfrac{1}{2}(9.81)(1.8)^2 = \tfrac{1}{2}(9.81)(3.24) = 15.9 \text{ m}$ .

Try this

Q1. State what the area under a velocity-time graph represents. [1 mark]

Cue. The displacement.

Q2. A stone is dropped from rest and falls for $2.0 \text{ s}$ . How far does it fall? [2 marks]

Cue. $s = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.81)(2.0)^2 = 19.6 \text{ m}$ .

Q3. State what the gradient of a displacement-time graph represents. [1 mark]

Cue. The velocity.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA car accelerates uniformly from

8.0 \text{ m s}^{-1}

20 \text{ m s}^{-1}

6.0 \text{ s}

. Calculate its acceleration and the distance travelled in this time.

Show worked answer →

Use the suvat equations with $u = 8.0$ , $v = 20$ , $t = 6.0$ .

Acceleration: $a = \dfrac{v - u}{t} = \dfrac{20 - 8.0}{6.0} = 2.0 \text{ m s}^{-2}$ .

Distance: $s = \dfrac{(u + v)}{2}t = \dfrac{(8.0 + 20)}{2} \times 6.0 = 14 \times 6.0 = 84 \text{ m}$ .

Markers reward selecting suitable suvat equations, correct substitution, and correct units. A common slip is to use $s = ut$ and forget the acceleration term.

AQA 20214 marksA ball is thrown vertically upwards with a speed of

15 \text{ m s}^{-1}

from the edge of a cliff. Calculate the maximum height it reaches above the launch point and the time taken to reach it. Take

g = 9.81 \text{ m s}^{-2}

Show worked answer →

Take upwards as positive, so $a = -9.81 \text{ m s}^{-2}$ . At the highest point the velocity is momentarily zero, $v = 0$ .

For the height, use $v^2 = u^2 + 2as$ : $0 = (15)^2 + 2(-9.81)s$ , so $s = \dfrac{225}{19.62} = 11.5 \text{ m}$ .

For the time, use $v = u + at$ : $0 = 15 - 9.81t$ , so $t = \dfrac{15}{9.81} = 1.5 \text{ s}$ .

Markers reward a clear sign convention, $v = 0$ at the top, and correct use of two suvat equations.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)