Forces and motion, Year 11: motion graphs and acceleration

speed

HintThe gradient is metres gained divided by seconds taken.

The whyGradient means the rise divided by the run. On these axes the rise is a distance and the run is a time, so the gradient is a distance divided by a time. A steeper line means the object is covering more metres in each second.

It is increasing

HintCompare how much distance is added in each second early on and later.

The whyThe gradient of a distance–time graph is the speed, so a line that gets steeper shows a speed that is rising: the car is accelerating. A straight sloping line would mean a steady speed, and a horizontal line would mean the car is not moving.

A straight line through the origin — the car moves at a steady speed

HintLook at how much distance is added in each two-second step.

The whyThe car gains the same 3 m in every 2 s, so the points lie on a straight line starting at the origin. An unchanging gradient means an unchanging speed; here it is 3 ÷ 2 = 1.5 m/s. Time goes on the horizontal axis and distance on the vertical axis.

4

HintFind how far the line rises and how far it runs between the two points.

The whySpeed = gradient = change in distance ÷ change in time = (130 − 50) ÷ (30 − 10) = 80 ÷ 20. Always use the changes between two points on the line, not the coordinates of a single point.

8 m/s

HintThe tangent is a straight line, so find its gradient in the usual way.

The whyFor a curve the gradient is different at every moment, so you draw the straight line that just touches the curve at the time you want and measure the gradient of that line: (32 − 0) ÷ (6 − 2) = 32 ÷ 4. A bigger triangle on the tangent gives a more accurate answer.

2 m/s²

HintFirst find how much her speed changed, then share that change across the time it took.

The whyAcceleration = change in velocity ÷ time taken, or a = Δv ÷ t; AQA expects this equation to be recalled. Her velocity rose by 6 m/s (from 2 to 8), and 6 ÷ 3 = 2, so she gained 2 m/s in every second. Only the change is divided by the time.

second

HintRead the unit aloud slowly and listen for the word you say twice.

The whym/s² is short for 'metres per second, per second': a speed (m/s) gained each second. A car accelerating at 3 m/s² from rest does 3 m/s after one second, 6 m/s after two and 9 m/s after three.

Car A

HintWork out how much speed each one gains in a single second.

The whyCar A gains 5 m/s every second (10 ÷ 2); Car B gains only 3 m/s every second (30 ÷ 10). Acceleration compares how quickly speed changes, not how much it changes altogether — B ends up faster simply because it kept going for longer.

Deceleration

HintIt is what a driver causes by pressing the brake pedal.

The whyIn physics slowing down is still an acceleration — a negative one, because the speed is falling rather than rising. The separate word is handy, but the idea underneath is the same: the velocity is changing, and a force is making it change.

a ~ 3 m/s²

HintDivide the change in velocity by the time, and do not claim more accuracy than the question gives.

The whyThe symbol ~ means 'is approximately'. Here a = Δv ÷ t = 27 ÷ 9, and because both figures are rough the answer is given as about 3 m/s². Estimating everyday accelerations like this is a GCSE skill: a few m/s² is typical for a car pulling away briskly.

acceleration

HintThe gradient here is metres per second gained, divided by seconds taken.

The whyOn these axes the rise is a change in velocity and the run is a time, so the gradient is change in velocity ÷ time. A steeper line means the velocity is changing more quickly, and a line sloping downwards means the object is slowing.

2

HintFind the change in velocity between the two points and the time it took.

The whyAcceleration = gradient = change in velocity ÷ time taken = (12 − 2) ÷ (5 − 0) = 10 ÷ 5. The line does not start at zero velocity, so it is the change of 10 m/s that matters, not the final 12 m/s.

It moves at a steady 8 m/s, with zero acceleration

HintRead the vertical axis label again before you answer.

The whyA horizontal line on these axes means the velocity is not changing: the gradient, and so the acceleration, is zero. The object is still moving, covering 8 m in every second. It would only be at rest if the line lay along the time axis, at 0 m/s.

−3 m/s² (a deceleration of 3 m/s²)

HintWork out final minus initial velocity, and keep the sign.

The whyAcceleration = change in velocity ÷ time = (6 − 18) ÷ 4 = −12 ÷ 4. The minus sign shows the velocity is falling: a line that slopes downwards has a negative gradient, and the object is decelerating.

The distance travelled (or the displacement)

HintMultiply the units of the two axes together and see what you get.

The whyHeight × width on this graph is a velocity × a time, which is a distance: metres per second × seconds gives metres. So the area under the line, found by splitting it into rectangles and triangles or by counting squares, is how far the object has gone.

80 m

HintSplit the shape under the line into a triangle and a rectangle.

The whyDistance is the area under the line. The triangle for the first 4 s has area ½ × 4 × 10 = 20 m. The rectangle from 4 s to 10 s has area 6 × 10 = 60 m. Adding the two gives the total distance.