AS & A Level · Practice questions · Variable acceleration: using calculus

Rates of change.

Six original Cambridge-style questions on kinematics with calculus, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Down by differentiating, up by integrating.

Down: \(v = \dfrac{ds}{dt}\), \(a = \dfrac{dv}{dt}\).
Up: integrate, then use a known value to find the constant.
At rest: \(v = 0\). Maximum velocity: \(a = 0\).
Distance: find where \(v = 0\), split, and add the sizes.
Stages: the velocity at the end of one formula starts the next.

Recall

[3 marks]

A particle moves so that \(s = 2t^3 - 3t^2 + 5\). Find expressions for \(v\) and \(a\), and their values when \(t = 2\).

\(v = 6t^2 - 6t\); at \(t = 2\), \(v = 24 - 12 = 12\ \text{m s}^{-1}\). ✓
\(a = 12t - 6\); at \(t = 2\), \(a = 18\ \text{m s}^{-2}\). ✓

Calculation

[4 marks]

A particle starts from O and moves with velocity \(v = 3\sqrt{t}\ \text{m s}^{-1}\). Find its acceleration when \(t = 4\) and its displacement from O at that time.

\(a = \tfrac{3}{2}t^{-1/2}\), so at \(t = 4\), \(a = \tfrac{3}{2} \times \tfrac{1}{2} = 0.75\ \text{m s}^{-2}\). ✓
\(s = \displaystyle\int_0^4 3t^{1/2}\,dt = \left[2t^{3/2}\right]_0^4 = 2 \times 8 = 16\) m. ✓

Calculation

[5 marks]

A particle starts at O with velocity \(3\ \text{m s}^{-1}\) and has acceleration \(a = (4 - 2t)\ \text{m s}^{-2}\). Find its maximum velocity and the time at which it is next at rest, giving the time in exact form.

\(v = 3 + 4t - t^2\). ✓
\(a = 0\) at \(t = 2\): maximum \(v = 7\ \text{m s}^{-1}\). ✓
\(v = 0\): \(t^2 - 4t - 3 = 0\), \(t = 2 \pm \sqrt{7}\). ✓
\(t > 0\), so \(t = 2 + \sqrt{7}\) (about 4.65 s). ✓

Analysis

[5 marks]

A particle moves so that \(s = t^3 - 6t^2 + 9t\) m. Find the times when it is at rest and the total distance it travels in the first 4 s.

\(v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)\): at rest at \(t = 1\) and \(t = 3\). ✓
\(s(0) = 0\), \(s(1) = 4\), \(s(3) = 0\), \(s(4) = 4\). ✓
Distance \(= 4 + 4 + 4 = 12\) m. ✓

Analysis

[5 marks]

A particle leaves O with velocity \(v = 8t - 2t^2\ \text{m s}^{-1}\) for \(0 \le t \le 6\). Show that it returns to O at \(t = 6\), and find the total distance travelled.

\(s = 4t^2 - \tfrac{2}{3}t^3\) (since \(s = 0\) at \(t = 0\)); \(s(6) = 144 - 144 = 0\), so it is back at O. ✓
\(v = 2t(4 - t) = 0\) at \(t = 4\): it turns round there. ✓
\(s(4) = 64 - \tfrac{128}{3} = \tfrac{64}{3}\). ✓
Distance \(= \tfrac{64}{3} + \tfrac{64}{3} = \tfrac{128}{3} = 42.7\) m. ✓

Analysis

[7 marks]

A particle P starts from rest at O. For \(0 \le t \le 4\) its velocity is \(v = 1.5t^2 - 0.25t^3\ \text{m s}^{-1}\). For \(t \ge 4\) it decelerates uniformly at \(2\ \text{m s}^{-2}\) until it comes to rest.

(a) Show that the speed of P at \(t = 4\) is \(8\ \text{m s}^{-1}\) and that its acceleration at that instant is zero.

(b) Find the total distance travelled by P.

(a) \(v(4) = 24 - 16 = 8\). \(a = 3t - 0.75t^2\), so \(a(4) = 12 - 12 = 0\). ✓
(b) Stage 1: \(\displaystyle\int_0^4 (1.5t^2 - 0.25t^3)\,dt = \left[0.5t^3 - \tfrac{1}{16}t^4\right]_0^4 = 32 - 16 = 16\) m. ✓
Stage 2: from 8 to 0 at \(2\ \text{m s}^{-2}\) takes 4 s, distance \(= \tfrac{1}{2}(8 + 0)(4) = 16\) m. ✓
Total distance \(= 32\) m. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.