Acceleration and Motion Graphs

When the lights turn green a car pulls away, its speed climbing from zero; when you brake at a junction, your speed falls back to zero. Any change of velocity — speeding up, slowing down or changing direction — is called acceleration, and it is one of the central ideas in describing motion. Two kinds of graph let us picture motion at a glance: the distance–time graph, whose gradient tells you the speed, and the velocity–time graph, whose gradient tells you the acceleration and whose area tells you the distance travelled. This lesson, part of Topic P2 (Forces) of OCR Gateway Science A, defines acceleration and its equation, then shows how to draw and read both kinds of graph.

By the end of this lesson you should be able to define acceleration and use the equation $a = \dfrac{\Delta v}{t}$a=tΔv​, interpret distance–time graphs (gradient = speed), and interpret velocity–time graphs (gradient = acceleration, area under = distance), including curved sections.

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What Is Acceleration?

Acceleration is the rate of change of velocity — how quickly the velocity changes each second. It is a vector (it has a direction). It is given by:

$a = \dfrac{\Delta v}{t}$a=tΔv​

where $a$a is the acceleration (in $\text{m/s}^2$m/s2), $\Delta v$Δv is the change in velocity (in $\text{m/s}$m/s) and $t$t is the time taken for that change (in $\text{s}$s). The symbol $\Delta v$Δv (delta vee) means "the change in velocity", always the final velocity minus the starting velocity:

$\Delta v = v - u$Δv=v−u

where $u$u is the starting (initial) velocity and $v$v is the final velocity.

The unit of acceleration is metres per second squared ($\text{m/s}^2$m/s2). An acceleration of $2 \text{ m/s}^2$2 m/s2 means the velocity increases by $2 \text{ m/s}$2 m/s every second.

A negative acceleration means the object is slowing down — this is sometimes called deceleration. If a car slows from $20 \text{ m/s}$20 m/s to rest, its change in velocity is $0 - 20 = -20 \text{ m/s}$0−20=−20 m/s, so the acceleration comes out negative.

Exam Tip: Acceleration is the change in velocity ÷ time, in $\text{m/s}^2$m/s2. Always work out $\Delta v$Δv as final − start; if the object slows down, $\Delta v$Δv is negative, giving a negative (decelerating) acceleration.

Worked Example 1 — Acceleration of a car

A car speeds up from $5 \text{ m/s}$5 m/s to $25 \text{ m/s}$25 m/s in $8 \text{ s}$8 s. Calculate its acceleration.

Step 1 — find the change in velocity: $\Delta v = v - u = 25 - 5 = 20 \text{ m/s}$Δv=v−u=25−5=20 m/s.

Step 2 — write the equation: $a = \dfrac{\Delta v}{t}$a=tΔv​.

Step 3 — substitute: $a = \dfrac{20}{8}$a=820​.

Step 4 — calculate: $a = 2.5 \text{ m/s}^2$a=2.5 m/s2.

Answer: the acceleration is $2.5 \text{ m/s}^2$2.5 m/s2.

Worked Example 2 — Deceleration

A cyclist slows from $12 \text{ m/s}$12 m/s to $4 \text{ m/s}$4 m/s in $4 \text{ s}$4 s. Calculate the acceleration.

Step 1 — find $\Delta v$Δv: $\Delta v = 4 - 12 = -8 \text{ m/s}$Δv=4−12=−8 m/s.

Step 2 — write the equation: $a = \dfrac{\Delta v}{t}$a=tΔv​.

Step 3 — substitute and calculate: $a = \dfrac{-8}{4} = -2 \text{ m/s}^2$a=4−8​=−2 m/s2.

Answer: the acceleration is $-2 \text{ m/s}^2$−2 m/s2. The minus sign shows the cyclist is decelerating (slowing down).

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Distance–Time Graphs

A distance–time graph plots the distance travelled (on the vertical axis) against the time taken (on the horizontal axis). The key rule is:

The gradient (steepness) of a distance–time graph equals the speed.

This follows directly from $\text{speed} = \dfrac{\text{distance}}{\text{time}}$speed=timedistance​, which is exactly "vertical change ÷ horizontal change", the definition of gradient. So:

• a horizontal line (zero gradient) means the object is stationary (distance not changing);
• a straight sloping line (constant gradient) means a constant speed;
• a steeper line means a higher speed;
• a curved line means a changing speed — curving upwards (getting steeper) means speeding up, and curving so it flattens means slowing down.

To find the speed from a straight section, calculate the gradient: pick two points on the line, and divide the change in distance (vertical) by the change in time (horizontal). For a curved line, the instantaneous speed at a point is the gradient of the tangent drawn to the curve at that point.

Worked Example 3 — Speed from a distance–time graph

On a distance–time graph, an object moves from $20 \text{ m}$20 m at $2 \text{ s}$2 s to $80 \text{ m}$80 m at $8 \text{ s}$8 s along a straight line. Calculate its speed.

Step 1 — speed is the gradient: $\text{speed} = \dfrac{\text{change in distance}}{\text{change in time}}$speed=change in timechange in distance​.

Step 2 — find the changes: change in distance $= 80 - 20 = 60 \text{ m}$=80−20=60 m; change in time $= 8 - 2 = 6 \text{ s}$=8−2=6 s.

Step 3 — calculate: $\text{speed} = \dfrac{60}{6} = 10 \text{ m/s}$speed=660​=10 m/s.

Answer: the speed is $10 \text{ m/s}$10 m/s.

Exam Tip: On a distance–time graph the gradient is the speed. A flat line means stopped; a steeper line means a faster speed. For a curve, draw a tangent and find its gradient to get the instantaneous speed.

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Velocity–Time Graphs

A velocity–time graph plots velocity (vertical axis) against time (horizontal axis). It carries two pieces of information, which makes it especially powerful:

1. The gradient of a velocity–time graph equals the acceleration. A steep upward slope is a large acceleration; a downward slope is a deceleration; a horizontal line is constant velocity (zero acceleration).
2. The area under a velocity–time graph equals the distance travelled.

Reading the graph above from left to right: the velocity rises steadily from zero (constant acceleration), then stays level (constant velocity), then falls back to zero (deceleration). The shaded area under the whole line is the total distance travelled.

Finding the acceleration from a straight section means finding its gradient (change in velocity ÷ change in time) — which is exactly the acceleration equation $a = \dfrac{\Delta v}{t}$a=tΔv​.

Finding the distance means finding the area under the line. For simple shapes you split the area into rectangles and triangles: a rectangle's area is $\text{base} \times \text{height}$base×height, and a triangle's area is $\dfrac{1}{2} \times \text{base} \times \text{height}$21​×base×height.

Worked Example 4 — Acceleration from a velocity–time graph

On a velocity–time graph, the velocity rises in a straight line from $0 \text{ m/s}$0 m/s at $0 \text{ s}$0 s to $30 \text{ m/s}$30 m/s at $12 \text{ s}$12 s. Find the acceleration.

Step 1 — acceleration is the gradient: $a = \dfrac{\Delta v}{t}$a=tΔv​.

Step 2 — substitute: $a = \dfrac{30 - 0}{12 - 0} = \dfrac{30}{12}$a=12−030−0​=1230​.