Every card in Forces and motion, Year 11: motion graphs and acceleration
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- On a distance–time graph, the gradient of the line gives the object's ____.
speed
HintThe gradient is metres gained divided by seconds taken.
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.
- On a distance–time graph for a car, the line is a curve that gets steeper and steeper. What is happening to the car's speed?
It is increasing
HintCompare how much distance is added in each second early on and later.
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 pupil times a toy car and records: 0 s, 0 m; 2 s, 3 m; 4 s, 6 m; 6 s, 9 m. When these are plotted as a distance–time graph, what shape is the line and what does it show about the car's motion?
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.
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.
- On a distance–time graph a straight section runs from the point (10 s, 50 m) to the point (30 s, 130 m). Calculate the speed during this section in m/s. (number only)
4
HintFind how far the line rises and how far it runs between the two points.
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.
- The distance–time graph for an accelerating car is a curve. To find the speed at 4 s, a tangent is drawn touching the curve at 4 s. The tangent passes through the points (2 s, 0 m) and (6 s, 32 m). What is the car's speed at 4 s?
8 m/s
HintThe tangent is a straight line, so find its gradient in the usual way.
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.
- A cyclist speeds up from 2 m/s to 8 m/s in 3 s. What is her acceleration?
2 m/s²
HintFirst find how much her speed changed, then share that change across the time it took.
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.
- An acceleration of 3 m/s² means the speed rises by 3 m/s in every ____.
second
HintRead the unit aloud slowly and listen for the word you say twice.
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 gains 10 m/s of speed in 2 s. Car B gains 30 m/s of speed in 10 s. Which car has the greater acceleration?
Car A
HintWork out how much speed each one gains in a single second.
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.
- An acceleration in the opposite direction to an object's motion, so the object slows down
Deceleration
HintIt is what a driver causes by pressing the brake pedal.
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 family car goes from rest to about 27 m/s (60 mph) in roughly 9 s. Write its approximate acceleration, using the symbol ~.
a ~ 3 m/s²
HintDivide the change in velocity by the time, and do not claim more accuracy than the question gives.
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.
- The gradient of a velocity–time graph gives the object's ____.
acceleration
HintThe gradient here is metres per second gained, divided by seconds taken.
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.
- On a velocity–time graph a straight line runs from the point (0 s, 2 m/s) to the point (5 s, 12 m/s). Calculate the acceleration in m/s². (number only)
2
HintFind the change in velocity between the two points and the time it took.
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.
- On a velocity–time graph, the line is horizontal at 8 m/s for ten seconds. Describe the object's motion during that time.
It moves at a steady 8 m/s, with zero acceleration
HintRead the vertical axis label again before you answer.
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.
- On a velocity–time graph a straight line falls from 18 m/s at 0 s to 6 m/s at 4 s. What is the acceleration?
−3 m/s² (a deceleration of 3 m/s²)
HintWork out final minus initial velocity, and keep the sign.
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.
- On a velocity–time graph, what does the area between the line and the time axis represent?
The distance travelled (or the displacement)
HintMultiply the units of the two axes together and see what you get.
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.
- A velocity–time graph rises in a straight line from 0 m/s at 0 s to 10 m/s at 4 s, then stays level at 10 m/s until 10 s. How far does the object travel in the whole 10 s?
80 m
HintSplit the shape under the line into a triangle and a rectangle.
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.
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