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How do you describe and calculate the motion of an object, using graphs and equations of constant acceleration?

Displacement, velocity and acceleration, motion graphs and the meaning of their gradients and areas, the constant acceleration equations, motion under gravity, and using calculus to relate displacement, velocity and acceleration.

A focused answer to the AQA A-Level Mathematics kinematics content, covering displacement, velocity and acceleration, motion graphs, the constant acceleration equations, motion under gravity, and using calculus for variable acceleration.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. Displacement, velocity and acceleration
  3. Motion graphs
  4. Constant acceleration equations
  5. Variable acceleration with calculus

What this dot point is asking

AQA wants you to describe motion using displacement, velocity and acceleration, interpret motion graphs through gradients and areas, apply the constant acceleration (suvat) equations, handle motion under gravity, and use calculus when acceleration varies. Kinematics spans both the constant-acceleration toolkit and the calculus link, and questions often combine a graph with an algebraic calculation.

Displacement, velocity and acceleration

These three quantities are vectors in one dimension, so they carry a sign indicating direction along the line. Displacement is position relative to a fixed origin, velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Distance and speed are the scalar magnitudes; when motion reverses, total distance exceeds the magnitude of net displacement, and average speed exceeds the magnitude of average velocity.

Motion graphs

Reading graphs is a reliable source of marks. A horizontal displacement-time line means the object is stationary; a straight sloping velocity-time line means constant acceleration. Splitting a velocity-time graph into triangles and rectangles is often the quickest way to find total distance, as in the trapezium example above.

Constant acceleration equations

Choosing the right equation is a matter of listing the five quantities, marking the three you know and the one you want, and picking the equation that omits the fifth. This avoids unnecessary algebra.

Variable acceleration with calculus

When acceleration is not constant the suvat equations no longer apply, and you use calculus instead. Differentiating displacement with respect to time gives velocity, and differentiating velocity gives acceleration: v=dsdtv = \frac{ds}{dt} and a=dvdta = \frac{dv}{dt}. Reversing this, you integrate acceleration to find velocity and integrate velocity to find displacement, fixing the constant of integration from given conditions (such as an initial position or velocity). For example, if s=t3−2ts = t^3 - 2t, then v=3t2−2v = 3t^2 - 2 and a=6ta = 6t.

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 20197 marksPaper 2, Section B. A particle moves in a straight line. It starts from rest and accelerates uniformly at 33 metres per second squared for 44 seconds. It then travels at constant velocity for 1010 seconds, before decelerating uniformly to rest in a further 66 seconds. (a) Sketch a velocity-time graph for the motion. (b) Find the maximum velocity. (c) Find the total distance travelled.
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For (a) the graph rises in a straight line from the origin, is horizontal in the middle, then falls back to zero, forming a trapezium. For (b) the maximum velocity comes from v=u+at=0+3×4=12v = u + at = 0 + 3 \times 4 = 12 metres per second. For (c) the distance is the area under the graph. The acceleration phase is a triangle of area 12×4×12=24\frac{1}{2} \times 4 \times 12 = 24 m; the constant phase is a rectangle of area 10×12=12010 \times 12 = 120 m; the deceleration phase is a triangle of area 12×6×12=36\frac{1}{2} \times 6 \times 12 = 36 m. Total distance =24+120+36=180= 24 + 120 + 36 = 180 metres. Markers reward a correctly shaped graph, the suvat value for the peak, and area calculation for the distance.

AQA 20216 marksPaper 2, Section B. A particle moves along a straight line so that its displacement from the origin, in metres, is s=t3−6t2+9ts = t^3 - 6t^2 + 9t for t≥0t \ge 0 seconds. (a) Find expressions for the velocity and the acceleration at time tt. (b) Find the times at which the particle is instantaneously at rest. (c) Find the acceleration at the later of these times and state whether the particle is speeding up or slowing down.
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For (a), differentiate: v=dsdt=3t2−12t+9v = \frac{ds}{dt} = 3t^2 - 12t + 9 and a=dvdt=6t−12a = \frac{dv}{dt} = 6t - 12. For (b), set v=0v = 0: 3t2−12t+9=3(t−1)(t−3)=03t^2 - 12t + 9 = 3(t-1)(t-3) = 0, so t=1t = 1 and t=3t = 3 seconds. For (c), at t=3t = 3, a=6(3)−12=6a = 6(3) - 12 = 6 metres per second squared. Just after t=3t = 3 the velocity becomes positive while the acceleration is positive, so the particle is speeding up. Markers reward correct differentiation, factorising to find both rest times, and a justified statement using the signs of vv and aa.

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