IGCSE · Practice questions · Kinematics with calculus

Rest, turn, add up.

Six original Cambridge-style questions on kinematics with calculus: velocity and acceleration from displacement, displacement from acceleration with starting conditions, exponential and trig motion, and total distance with the turning points split out.

Original questions All questions on this page are original work, written in the Cambridge IGCSE 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: \(v = \dfrac{\mathrm{d}s}{\mathrm{d}t}\), \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t}\).
Up: \(v = \int a\,\mathrm{d}t\), \(s = \int v\,\mathrm{d}t\), and fix each constant from the given condition at once.
Key words: initially \(t = 0\); at rest \(v = 0\); at \(O\) \(s = 0\); maximum velocity \(a = 0\).
Distance: solve \(v = 0\), split the interval at each root, integrate each piece and add the sizes.
Graphs: gradient of \(s\)-\(t\) is \(v\), gradient of \(v\)-\(t\) is \(a\), area under \(v\)-\(t\) is displacement; speed is \(|v|\).

Recall

[2 marks]

A particle moves in a straight line so that its displacement from \(O\) after \(t\) seconds is \(s = 2t^{3} - 9t^{2} + 12t + 1\) metres. Find its velocity and its acceleration when \(t = 3\).

\(v = 6t^{2} - 18t + 12\), so \(v(3) = 54 - 54 + 12 = 12\) m s^-1. ✓
\(a = 12t - 18\), so \(a(3) = 36 - 18 = 18\) m s^-2. ✓

Calculation

[4 marks]

A particle moves in a straight line with velocity \(v = 3t^{2} - 21t + 30\) m s^-1 at time \(t\) seconds.

(a) Find the acceleration when \(t = 1\).

(b) Find the times when the particle is at rest, and its acceleration at each of these times.

(a) \(a = 6t - 21\), so \(a(1) = -15\) m s^-2. ✓
(b) \(3(t^{2} - 7t + 10) = 0\), so \(3(t - 2)(t - 5) = 0\) and \(t = 2\) or \(t = 5\). ✓
\(a(2) = 12 - 21 = -9\) m s^-2 and \(a(5) = 30 - 21 = 9\) m s^-2. ✓

Calculation

[4 marks]

A particle passes through \(O\) at time \(t = 0\) with velocity 4 m s^-1. Its acceleration after \(t\) seconds is \(a = 12t - 6\) m s^-2.

(a) Find \(v\) in terms of \(t\) and show that the particle is never at rest.

(b) Find the displacement of the particle from \(O\) when \(t = 2\).

(a) \(v = 6t^{2} - 6t + c\), and \(v(0) = 4\) gives \(v = 6t^{2} - 6t + 4\). ✓
For \(6t^{2} - 6t + 4 = 0\) the discriminant is \(36 - 96 = -60 \lt 0\), so \(v\) is never zero: the particle is never at rest. ✓
(b) \(s = 2t^{3} - 3t^{2} + 4t + k\), and \(s(0) = 0\) gives \(k = 0\). ✓
\(s(2) = 16 - 12 + 8 = 12\) m. ✓

Calculation

[5 marks]

A particle starts from \(O\) and moves in a straight line so that its velocity after \(t\) seconds is \(v = 8 - 8e^{-0.5t}\) m s^-1.

(a) Find the initial acceleration.

(b) Find an expression for the displacement \(s\) in terms of \(t\).

(d) Explain why the velocity never reaches 8 m s^-1.

(a) \(a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = 4e^{-0.5t}\), so \(a(0) = 4\) m s^-2. ✓
(b) \(s = 8t + 16e^{-0.5t} + c\). At \(t = 0\): \(0 = 16 + c\), so \(c = -16\) and \(s = 8t + 16e^{-0.5t} - 16\). ✓
(c) \(s(4) = 32 + 16e^{-2} - 16 = 16 + 16e^{-2} \approx 18.2\) m. ✓
(d) \(8e^{-0.5t} \gt 0\) for every \(t\), so \(v = 8 - 8e^{-0.5t}\) is always less than 8 (it only approaches 8 as \(t\) becomes large). ✓

Analysis

[5 marks]

A particle moves in a straight line so that its velocity after \(t\) seconds is \(v = 3\cos 2t\) m s^-1, and it is at \(O\) when \(t = 0\).

(a) Find \(s\) and \(a\) in terms of \(t\).

(b) Find the time when the particle first comes to rest, and its displacement and acceleration at that time.

(a) \(s = \dfrac{3}{2}\sin 2t + c\), and \(s(0) = 0\) gives \(c = 0\): \(s = \dfrac{3}{2}\sin 2t\). \(a = -6\sin 2t\). ✓
(b) \(\cos 2t = 0\) first when \(2t = \dfrac{\pi}{2}\), so \(t = \dfrac{\pi}{4}\). ✓
Then \(s = \dfrac{3}{2}\sin \dfrac{\pi}{2} = 1.5\) m and \(a = -6\sin \dfrac{\pi}{2} = -6\) m s^-2. ✓
(c) The particle moves out to \(s = 1.5\) at \(t = \dfrac{\pi}{4}\), then back: \(s\left(\dfrac{\pi}{2}\right) = \dfrac{3}{2}\sin \pi = 0\). ✓
Distance \(= 1.5 + 1.5 = 3\) m (the displacement is 0). ✓

Analysis

[6 marks]

A particle moves in a straight line so that its displacement from \(O\) after \(t\) seconds is \(s = t^{3} - 9t^{2} + 15t\) metres, for \(0 \le t \le 6\).

(a) Find the times when the particle is at rest.

(b) Find the total distance travelled.

(a) \(v = 3t^{2} - 18t + 15 = 3(t - 1)(t - 5) = 0\), so \(t = 1\) and \(t = 5\). ✓
(b) \(s(0) = 0\), \(s(1) = 1 - 9 + 15 = 7\), \(s(5) = 125 - 225 + 75 = -25\), \(s(6) = 216 - 324 + 90 = -18\). ✓
Distance \(= 7 + (7 - (-25)) + (-18 - (-25)) = 7 + 32 + 7 = 46\) m. ✓
(c) \(a = 6t - 18\) is zero at \(t = 3\), where \(v = 27 - 54 + 15 = -12\). The ends give \(v(0) = 15\) and \(v(6) = 108 - 108 + 15 = 15\). ✓
Speed \(= |v|\): 15 at \(t = 0\), down to 0 at \(t = 1\), up to 12 at \(t = 3\), down to 0 at \(t = 5\), up to 15 at \(t = 6\). Greatest speed 15 m s^-1 (at \(t = 0\) and \(t = 6\)). ✓
The speed-time graph is the \(v\)-\(t\) parabola with the part between \(t = 1\) and \(t = 5\) reflected above the axis, so it touches zero twice. ✓

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.