Speed from a Curved Travel Graph

Draw a tangent to a curved distance-time or speed-time graph to estimate the speed or acceleration at one moment.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1 · 13 real past-paper questions below

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Speed at One Moment

On a curved travel graph the steepness keeps changing, so the speed keeps changing too.

To find it at one moment, draw a tangent: a straight line that just touches the curve at that point. Its gradient is the answer.

Graph	Tangent's gradient gives
Distance–time	speed at that moment
Speed–time	acceleration at that moment

Rule

Tangent with a ruler, two points on it far apart, then up ÷ across.

Speed From a Distance–Time Curve

The 3-mark question: “Estimate the speed at 4 seconds.” Draw the tangent at $t = 4$ , then find its gradient.

Worked example

1
$t = 4 : d = 8 m$ — Go up from 4 on the time axis to the curve, then across to the distance axis. The point is (4, 8).
2
$tangent at (4, 8)$ — Lay a ruler so it just touches the curve at (4, 8) and does not cut across it, then draw the straight line.
3
$(2, 0) and (6, 16)$ — Read two points on your tangent where it crosses grid lines. Pick them far apart, so a small reading slip matters less.
4
$up = 16 - 0 = 16, across = 6 - 2 = 4$ — Up is the change in distance and across is the change in time. Subtract the smaller reading from the larger.
5
$\frac{16}{4} = 4$ — The gradient of a line is up ÷ across.
6
$4 m/s$ — Metres divided by seconds gives m/s. This is the speed at 4 seconds.

Common mistake

Reading two points on the curve. That is not the tangent, so the gradient is wrong.

Key idea

Leave the tangent and the two points you read on the graph. Marks are for the ruled tangent and for the answer, and an answer inside the accepted range scores even though it is an estimate.

Acceleration From a Speed–Time Curve

The same method on a speed–time graph gives the acceleration. A tangent going down means slowing down.

Worked example

1
$t = 4 : v = 12 m/s$ — Go up from 4 on the time axis to the curve, then across to the speed axis. The point is (4, 12).
2
$tangent at (4, 12)$ — Lay a ruler so it just touches the curve at (4, 12) and draw the straight line.
3
$(2, 20) and (7, 0)$ — Read two points on your tangent, far apart.
4
$up = 0 - 20 = - 20, across = 7 - 2 = 5$ — The tangent goes down as it moves right, so “up” is negative.
5
$\frac{- 20}{5} = - 4$ — The gradient is up ÷ across.
6
$- 4 m/s^{2}$ — Speed in m/s over time in s gives m/s². The minus sign means the speed falls by 4 m/s every second.

Common mistake

Dropping the minus sign. If the tangent goes down, the acceleration is negative.

Average Speed Is Different

“Average speed over the first 6 seconds” is not a tangent. Join the start and end points of that time.

Worked example

1
$t = 0 : d = 0$ — Read the distance at the start of the time.
2
$t = 6 : d = 18$ — Read the distance at the end of the time. The point is (6, 18).
3
$distance = 18 - 0 = 18 m, time = 6 - 0 = 6 s$ — Take the start from the end to get the total distance and the total time.
4
$\frac{18}{6} = 3 m/s$ — Average speed is total distance ÷ total time.

Key idea

On the same curve the speed at 4 seconds was 4 m/s. The average hides how fast it was going at one moment.

Which Method? Read the Question

Check the graph type and whether the question says “at” one time or “over” a time.

You see	You do
Speed at a time, curved distance–time	tangent, up ÷ across
Acceleration at a time, curved speed–time	tangent, up ÷ across
Average speed over a time	total distance ÷ total time
When is it fastest?	where the curve is steepest
Straight-line graph	Distance-Time Graphs or Speed-Time Graphs and Acceleration
An equation, exact gradient	Gradient of a Curve at a Point

Key idea

Your tangent is drawn by hand, so the answer is an estimate. Mark schemes accept a range.

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