Exam-style questionsVariable Acceleration questions with mark schemes
Easiest first. Try each question on paper, then open its mark scheme: M marks are for method, A marks for accurate answers that follow from the method, B marks for independent results.
Question 1easy4 marksYear 12
A particle moves along a straight line. At time \(t\) seconds its velocity is \(v\) m s\(^{-1}\), where \(v = 3 t^{2} - 12 t + 9\), \(t \ge 0\).
(a) Find the values of \(t\) when the particle is instantaneously at rest. (2)
(b) Find the acceleration of the particle when \(t = 4\). (2)
Mark scheme and answer
(a) \(3(t-1)(t-3) = 0\) M1
\(t = 1\) and \(t = 3\) A1
(b) \(a = \dfrac{dv}{dt} = 6 t - 12\) M1
\(a = 12\) m s\(^{-2}\) A1
Answer: (a) t = 1, 3; (b) 12 m s⁻²
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 2easy5 marksYear 12
A particle moves in a straight line such that its displacement from \(O\) after \(t\) seconds is \(s = 5t + 2t^2 - \tfrac13t^3\) m.
(a) Find the initial velocity. (2)
(b) Find the time at which the particle comes to instantaneous rest. (2)
(c) Find the acceleration at this time. (1)
Mark scheme and answer
(a) \(v = - t^{2} + 4 t + 5\) M1
\(v(0) = 5\) m s\(^{-1}\) A1
(b) \(t^2 - 4t - 5 = 0\) M1
\(t = 5\) s A1
(c) \(a = 4 - 2 t\); \(a(5) = -6\) m s\(^{-2}\) B1
Answer: (a) 5 m s⁻¹; (b) t = 5; (c) −6 m s⁻²
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 3medium5 marksYear 12
A cyclist moves along a straight road. At time \(t\) seconds her acceleration is \((6 t - 4)\) m s\(^{-2}\). At \(t = 0\) she passes point \(O\) with velocity 2 m s\(^{-1}\). (The model applies for \(0 \le t \le 3\).)
(a) Find an expression for her velocity at time \(t\). (2)
(b) Find her distance from \(O\) when \(t = 3\). (3)
Mark scheme and answer
(a) \(v = \int (6t-4)\,dt = 3t^2 - 4t + c\) M1
\(v = 3 t^{2} - 4 t + 2\) A1
(b) \(s = t^{3} - 2 t^{2} + 2 t\) (+0) M1A1
\(s(3) = 15\) m A1
Answer: (a) v = 3t^2 − 4t + 2; (b) 15 m
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\). Check v > 0 on [0,3] (minimum 2/3), so displacement = distance.
Question 4medium6 marksYear 12
A lift moves vertically upwards from rest. Its velocity after \(t\) seconds is \(v = - 2 t^{2} + 8 t\) m s\(^{-1}\) for \(0 \le t \le 4\).
(a) Find the maximum velocity of the lift. (3)
(b) Find the distance travelled by the lift in the 4 seconds. (2)
(c) Explain why the model is only valid for \(0 \le t \le 4\). (1)
Mark scheme and answer
(a) \(\dfrac{dv}{dt} = 8 - 4t = 0\), \(t = 2\) M1A1
\(v_{max} = 8\) m s\(^{-1}\) A1
(b) \(\int_0^4 (8t - 2t^2)\,dt\) M1
\(= \frac{64}{3}\) m (\(= 21.3\)) A1
(c) \(v = 0\) at \(t = 4\); for \(t \gt 4\) the model gives negative velocity (the lift would go down), which is not intended B1
Answer: (a) 8 m s⁻¹; (b) 64/3 m; (c) v < 0 for t > 4
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 5medium6 marksYear 12
A particle moves along the \(x\)-axis. Its displacement from \(O\) at time \(t\) seconds is \(x = 2 t^{3} - 15 t^{2} + 24 t + 5\) metres.
(a) Find the values of \(t\) for which the particle is at rest. (3)
(b) Find the positions of the particle at these times. (2)
(c) Find the acceleration when \(t = 2\). (1)
Mark scheme and answer
(a) \(v = 6 t^{2} - 30 t + 24 = 6(t-1)(t-4)\) M1A1
\(t = 1,\ t = 4\) A1
(b) \(x(1) = 16\), \(x(4) = -11\) M1A1
(c) \(a = 12 t - 30\); \(a(2) = -6\) m s\(^{-2}\) B1
Answer: (a) t = 1, 4; (b) x = 16 m, −11 m; (c) −6 m s⁻²
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 6medium5 marksYear 12
A remote-controlled car moves along a straight path. Its velocity after \(t\) seconds is \(v = 6\sqrt t - t\) m s\(^{-1}\), for \(0 \le t \le 16\).
(a) Find the maximum velocity of the car. (3)
(b) Find the distance travelled in the first 9 seconds. (2)
Mark scheme and answer
(a) \(\dfrac{dv}{dt} = 3t^{-1/2} - 1 = 0\) M1
\(t = 9\) A1
\(v = 9\) m s\(^{-1}\) A1
(b) \(\int_0^9 (6t^{1/2} - t)\,dt = [4t^{3/2} - \tfrac12t^2]_0^9\) M1
\(= \frac{135}{2}\) m A1
Answer: (a) 9 m s⁻¹; (b) 135/2 m
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 7medium8 marksYear 13
A particle \(P\) moves in a plane. At time \(t\) seconds its position vector is \(\mathbf{r} = (t^{3} - 3 t)\mathbf{i} + (4 t^{2} - 2)\mathbf{j}\) metres.
(a) Find the velocity of \(P\) at time \(t\). (2)
(b) Find the speed of \(P\) when \(t = 2\). (2)
(c) Find the value of \(t\), \(t \gt 0\), when \(P\) is moving parallel to \(\mathbf{j}\). (2)
(d) Find the acceleration of \(P\) when \(t = 1\). (2)
Mark scheme and answer
(a) \(\mathbf{v} = (3 t^{2} - 3)\mathbf{i} + (8 t)\mathbf{j}\) M1A1
(b) \(\mathbf{v} = 9\mathbf{i} + 16\mathbf{j}\) M1
speed \(= \sqrt{81 + 256} = 18.4\) m s\(^{-1}\) A1
(c) \(3t^2 - 3 = 0\) M1
\(t = 1\) A1
(d) \(\mathbf{a} = (6 t)\mathbf{i} + (8)\mathbf{j}\) M1
\(= 6\mathbf{i} + 8\mathbf{j}\) m s\(^{-2}\) A1
Answer: (a) v = (3t^2 − 3)i + (8t)j; (b) √337 ≈ 18.4 m s⁻¹; (c) t = 1; (d) 6i + 8j m s⁻²
Board notes: Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 8medium4 marksYear 13
A particle of mass 0.5 kg moves under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds its position vector is \(\mathbf{r} = (2 t^{3} - t)\mathbf{i} + (- t^{4} + 5 t)\mathbf{j}\) m.
(a) Find the acceleration at time \(t\). (2)
(b) Find the magnitude of \(\mathbf{F}\) when \(t = 1\). (2)
Mark scheme and answer
(a) \(\mathbf{v} = (6t^2 - 1)\mathbf{i} + (5 - 4t^3)\mathbf{j}\) M1
\(\mathbf{a} = (12 t)\mathbf{i} + (- 12 t^{2})\mathbf{j}\) A1
(b) \(\mathbf{F} = 0.5(12\mathbf{i} -12\mathbf{j}) = 6\mathbf{i} -6\mathbf{j}\) M1
\(|\mathbf{F}| = 6 \sqrt{2} = 8.49\) N A1
Answer: (a) (12t)i + (−12t^2)j; (b) 8.49 N
Board notes: Links calculus with \(\mathbf{F} = m\mathbf{a}\). Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 9medium4 marksYear 13
A particle moves so that its position vector at time \(t\) seconds is \(\mathbf{r} = (t^{3})\mathbf{i} + (- t^{2} + 5 t)\mathbf{j}\) m.
(a) Find the acceleration of the particle when \(t = 1\). (2)
(b) Find the angle between the acceleration at \(t = 1\) and the vector \(\mathbf{i}\). (2)
Mark scheme and answer
(a) \(\mathbf{v} = 3t^2\mathbf{i} + (5 - 2t)\mathbf{j}\) M1
\(\mathbf{a} = 6t\mathbf{i} - 2\mathbf{j}\); at \(t = 1\): \(6\mathbf{i} -2\mathbf{j}\) A1
(b) \(\tan\theta = \tfrac26\) M1
\(\theta = 18.4°\) (below \(\mathbf{i}\)) A1
Answer: (a) 6i −2j m s⁻²; (b) 18.4°
Board notes: Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 10hard6 marksYear 12
A particle \(P\) moves on a straight line so that its displacement from a fixed point \(O\) at time \(t\) seconds is \(s\) metres, where \(s = t^{3} - 6 t^{2} + 9 t\).
(a) Find the times when \(P\) is at rest. (2)
(b) Find the total distance travelled by \(P\) in the first 4 seconds. (4)
Mark scheme and answer
(a) \(v = 3 t^{2} - 12 t + 9 = 0\) M1
\(t = 1,\ t = 3\) A1
(b) \(s(0)=0,\ s(1)=4,\ s(3)=0,\ s(4)=4\) M1A1
Distance \(= 4 + 4 + 4 = 12\) m M1A1
Answer: (a) t = 1, 3; (b) 12 m
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 11hard6 marksYear 12
A toy train moves along a straight track, starting at \(t = 0\). For \(0 \le t \le 3\) its velocity is \(v = 4 + 3t - t^2\) m s\(^{-1}\). At \(t = 3\) the power is cut and the train decelerates uniformly at 2 m s\(^{-2}\) until it stops.
(a) Find the distance travelled for \(0 \le t \le 3\). (2)
(b) Find the total distance travelled by the train. (4)
Mark scheme and answer
(a) \(\int_0^3 (4 + 3t - t^2)\,dt\) M1
\(= 12 + 13.5 - 9 = \frac{33}{2}\) m A1
(b) \(v(3) = 4\) B1
\(0 = 4^2 - 2(2)s\), \(s = 4\) M1A1
Total \(= \frac{41}{2}\) m A1
Answer: (a) 33/2 m; (b) 41/2 m
Board notes: Combines calculus with constant-acceleration (suvat) formulae. All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 12hard6 marksYear 12
A particle moves in a straight line from a point \(O\). At time \(t\) seconds its velocity is \(v = kt^2 - 5t\) m s\(^{-1}\), where \(k\) is a constant. When \(t = 3\), the particle has displacement 4.5 m from \(O\).
(a) Find the value of \(k\). (4)
(b) Find the acceleration of the particle when \(t = 3\). (2)
Mark scheme and answer
(a) \(s = \tfrac{k}{3}t^3 - \tfrac52t^2\) M1A1
\(9k - 22.5 = 4.5\) M1
\(k = 3\) A1
(b) \(a = 6 t - 5\), \(a(3) = 13\) m s\(^{-2}\) M1A1
Answer: (a) k = 3; (b) 13 m s⁻²
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 13hard7 marksYear 12
A particle starts from rest at \(O\) and moves in a straight line with acceleration \((12 - 6 t)\) m s\(^{-2}\) at time \(t\) seconds.
(a) Find expressions for the velocity and displacement at time \(t\). (3)
(b) Find the time at which the particle returns to \(O\). (2)
(c) Find the greatest distance of the particle from \(O\) before it returns. (2)
Mark scheme and answer
(a) \(v = - 3 t^{2} + 12 t\) M1A1
\(s = - t^{3} + 6 t^{2}\) A1
(b) \(6t^2 - t^3 = 0\) M1
\(t = 6\) s A1
(c) \(v = 0\) at \(t = 4\) M1
\(s(4) = 32\) m A1
Answer: (a) v = −3t^2 + 12t, s = −t^3 + 6t^2; (b) t = 6; (c) 32 m
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 14hard7 marksYear 13
A particle moves so that its velocity at time \(t\) seconds is \(\mathbf{v} = (6 t - 5)\mathbf{i} + (t^{2} - 4)\mathbf{j}\) m s\(^{-1}\). When \(t = 0\) the particle is at the point with position vector \((2\mathbf{i} - \mathbf{j})\) m.
(a) Find the position vector of the particle when \(t = 3\). (4)
(b) Find the values of \(t\) when the particle is moving parallel to the vector \(\mathbf{i} + \mathbf{j}\) (in either sense). (3)
Mark scheme and answer
(a) \(\mathbf{r} = (3t^2 - 5t + c_1)\mathbf{i} + (\tfrac13t^3 - 4t + c_2)\mathbf{j}\) M1A1
\(c_1 = 2,\ c_2 = -1\) M1
\(\mathbf{r}(3) = 14\mathbf{i} + (-4)\mathbf{j}\) m A1
(b) \(6t - 5 = t^2 - 4\) M1
\(t^2 - 6t + 1 = 0\), \(t = 3 \pm 2\sqrt2\) A1
\(t = 0.172\) or \(t = 5.83\); both valid (at the smaller value both components equal -3.97, so the motion is in the direction of \(-(\mathbf{i}+\mathbf{j})\), which is still parallel) A1
Answer: (a) 14i − 4j m; (b) t = 3 ± 2√2 (0.172 s and 5.83 s)
Board notes: Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 15hard7 marksYear 13
A particle \(P\) moves in a horizontal plane with acceleration \((2\mathbf{i} - 6t\mathbf{j})\) m s\(^{-2}\) at time \(t\) seconds, where \(\mathbf{i}\) points east and \(\mathbf{j}\) points north. At \(t = 0\), \(P\) is at the origin \(O\) with velocity \((3\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\).
(a) Find the velocity of \(P\) at time \(t\). (2)
(b) Find the position vector of \(P\) at time \(t\). (2)
(c) Find the distance \(OP\) when \(P\) is due east of \(O\) (\(t \gt 0\)). (3)
Mark scheme and answer
(a) \(\mathbf{v} = (2 t + 3)\mathbf{i} + (4 - 3 t^{2})\mathbf{j}\) M1A1
(b) \(\mathbf{r} = (t^{2} + 3 t)\mathbf{i} + (- t^{3} + 4 t)\mathbf{j}\) M1A1
(c) \(4t - t^3 = 0\), \(t = 2\) M1A1
\(x = 10\), so \(OP = 10\) m A1
Answer: (a) (2t + 3)i + (4 − 3t^2)j; (b) (t^2 + 3t)i + (−t^3 + 4t)j; (c) 10 m
Board notes: Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 16hard7 marksYear 13
A boat switches off its engine and coasts in a straight line. Its velocity \(t\) seconds later is modelled by \(v = 20e^{-0.2t}\) m s\(^{-1}\).
(a) Find the deceleration of the boat at the moment the engine is switched off. (2)
(b) Find the distance travelled in the first 5 seconds. (3)
(c) Show that, according to the model, the boat never travels more than 100 m, and comment on this. (2)
Mark scheme and answer
(a) \(\dfrac{dv}{dt} = -4e^{-0.2t}\) M1
At \(t = 0\): deceleration 4 m s\(^{-2}\) A1
(b) \(\int_0^5 20e^{-0.2t}\,dt = [-100e^{-0.2t}]_0^5\) M1A1
\(= 100(1 - e^{-1}) = 63.2\) m A1
(c) \(s = 100(1 - e^{-0.2t}) \lt 100\) for all \(t\) B1
Model predicts the boat never stops (v > 0 always), which is unrealistic; it will stop in finite time B1
Answer: (a) 4 m s⁻²; (b) 63.2 m; (c) s → 100
Board notes: Uses Year 2 calculus of \(e^{kx}\); AQA and OCR set similar "limiting distance" questions. All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 17hard7 marksYear 13
A piston moves in a straight line. Its velocity at time \(t\) seconds is \(v = 2\cos\left(\dfrac{\pi t}{3}\right)\) m s\(^{-1}\).
(a) Find the displacement of the piston from its starting position when \(t = 3\). (3)
(b) Find the total distance travelled by the piston in the interval \(0 \le t \le 3\). (2)
(c) Find the maximum magnitude of the acceleration. (2)
Mark scheme and answer
(a) \(s = \dfrac{6}{\pi}\sin\left(\dfrac{\pi t}{3}\right)\) M1A1
\(s(3) = 0\) A1
(b) \(v = 0\) at \(t = 1.5\) M1
Distance \(= 2\times\dfrac{6}{\pi} = \dfrac{12}{\pi} = 3.82\) m A1
(c) \(a = -\dfrac{2\pi}{3}\sin\left(\dfrac{\pi t}{3}\right)\) M1
max \(|a| = \dfrac{2\pi}{3} \approx 2.09\) m s\(^{-2}\) A1
Answer: (a) 0; (b) 12/π ≈ 3.82 m; (c) 2π/3 ≈ 2.09 m s⁻²
Board notes: Uses radians and Year 2 calculus of trigonometric functions. All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 18challenging7 marksYear 12
A particle moves in a straight line. At time \(t\) seconds its velocity is \(v = t^{2} - 5 t + 4\) m s\(^{-1}\). At \(t = 0\) it is at the point \(O\).
(a) Find the displacement of the particle from \(O\) when \(t = 5\). (3)
(b) Find the total distance travelled in the first 5 seconds. (4)
Mark scheme and answer
(a) \(s = \tfrac13t^3 - \tfrac52t^2 + 4t\) M1A1
\(s(5) = - \frac{5}{6}\) m A1
(b) \(v = 0\) at \(t = 1, 4\) M1
\(|s(1) - s(0)| = \tfrac{11}{6}\), \(|s(4) - s(1)| = \tfrac92\), \(|s(5) - s(4)| = \tfrac{11}{6}\) M1A1
Total \(= \frac{49}{6}\) m A1
Answer: (a) −5/6 m; (b) 49/6 m
Board notes: All boards: distance travelled must account for changes of direction (split the integral where \(v = 0\)). OCR may write \(\dot x\) and \(\ddot x\).
Question 19challenging9 marksYear 13
Two particles \(P\) and \(Q\) move in a plane. At time \(t\) seconds their position vectors are \(\mathbf{r}_P = t^2\mathbf{i} + 3t\mathbf{j}\) and \(\mathbf{r}_Q = (4t - 4)\mathbf{i} + (t + 4)\mathbf{j}\) metres.
(a) Show that \(P\) and \(Q\) collide and find the time of collision. (3)
(b) Find the distance between the particles at \(t = 0\). (2)
(c) Find the angle between the directions of motion of \(P\) and \(Q\) at the moment of collision. (4)
Mark scheme and answer
(a) \(t^2 = 4t - 4 \Rightarrow (t-2)^2 = 0\), \(t = 2\) M1A1
\(3t = t + 4 \Rightarrow t = 2\); same time for both components, so they collide at \(t = 2\) A1
(b) \(\mathbf{r}_P - \mathbf{r}_Q = 4\mathbf{i} - 4\mathbf{j}\) M1
distance \(= 4 \sqrt{2}\) m A1
(c) \(\mathbf{v}_P = 4\mathbf{i} + 3\mathbf{j},\ \mathbf{v}_Q = 4\mathbf{i} + 1\mathbf{j}\) M1A1
Angles with \(\mathbf{i}\): \(\tan^{-1}\tfrac34 = 36.87°\) and \(\tan^{-1}\tfrac14 = 14.04°\) M1
\(\theta = 36.87° - 14.04° = 22.8°\) A1
Answer: (a) t = 2; (b) 4√2 ≈ 5.66 m; (c) 22.8°
Board notes: The scalar product is not in A level Mathematics (it is Further Mathematics), so the scheme finds each direction as an angle with \(\mathbf{i}\); a correct scalar-product method is also credited. Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Question 20challenging6 marksYear 13
The velocity of a particle at time \(t\) seconds is \(\mathbf{v} = (2 t - 4)\mathbf{i} + (3 - t)\mathbf{j}\) m s\(^{-1}\).
(a) Show that the speed \(V\) satisfies \(V^2 = 5t^2 - 22t + 25\). (2)
(b) Find the minimum speed of the particle and the time at which it occurs. (3)
(c) Explain why the particle is never at rest. (1)
Mark scheme and answer
(a) \(V^2 = (2t-4)^2 + (3-t)^2\) M1
\(= 5 t^{2} - 22 t + 25\) A1
(b) \(\dfrac{d(V^2)}{dt} = 10t - 22 = 0\) M1
\(t = 11/5\) A1
\(V_{min} = \frac{2 \sqrt{5}}{5} = 0.894\) m s\(^{-1}\) A1
(c) The components are zero at different times (\(t = 2\) and \(t = 3\)), so \(\mathbf{v} \ne \mathbf{0}\); equivalently \(V_{min} \gt 0\) B1
Answer: (a) shown; (b) 0.894 m s⁻¹ at t = 11/5; (c) components never simultaneously zero
Board notes: Completing the square is equally valid. Vectors may be written in \(\mathbf{i},\mathbf{j}\) form or as column vectors; AQA and OCR frequently use column vectors.
Original questions in the style of Edexcel 9MA0, independently checked before publishing. Not past-paper questions. Covers: Differentiating velocity, Differentiating displacement, Integrating acceleration, Maximum velocity, Calculus with fractional powers, Vectors: differentiating position, Vectors: force from position, Vectors: direction of acceleration, Distance travelled with change of direction, Piecewise motion, Finding constants, Vectors: integrating velocity, Vectors: integrating acceleration, Calculus with exponentials, Calculus with trigonometric functions, Vectors: two particles, Vectors: minimum speed.
A Level Math Revision