Variable acceleration in a straight line - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 18 exam-style questions on Variable acceleration in a straight line for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
3–9 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A particle \(P\) moves along a straight line, starting from a fixed point \(O\) at time \(t = 0\). At time \(t\) seconds, the velocity of \(P\) is \(v\ \text{m s}^{-1}\), where
\[v = 3 + 2t - 0.25t^2, \quad 0 \leqslant t \leqslant 5\]
Show that the distance travelled by \(P\) during the fourth second of its motion is \(\dfrac{83}{12}\) m.

Model answer

\(v > 0\) for \(3 \leqslant t \leqslant 4\) (\(v(3) = 6.75\), \(v(4) = 7\)), so the distance is the definite integral.
The fourth second is from \(t = 3\) to \(t = 4\).
\(\displaystyle\int_3^4 (3 + 2t - 0.25t^2)\,dt = \left[3t + t^2 - \dfrac{t^3}{12}\right]_3^4\)
\(= \left(12 + 16 - \dfrac{64}{12}\right) - \left(9 + 9 - \dfrac{27}{12}\right) = \dfrac{272}{12} - \dfrac{189}{12} = \dfrac{83}{12}\) m

Explanation

The fourth second of the motion runs from \(t = 3\) to \(t = 4\), so integrate the velocity between these limits. The particle does not change direction in this interval, so the integral is the distance. The common error is to use the limits 0 and 4, which gives the distance in the first four seconds.

Mark scheme

1 mark
M1 Integrates \(v\), with at least one power increased by 1. An answer with no working scores 0 marks
An answer with no working scores 0 marks (show that).
1 mark
M1 (dM1) Uses the limits \(t = 3\) and \(t = 4\) (the fourth second): \(\left[3t + t^2 - \dfrac{t^3}{12}\right]_3^4\)
A common error is to use the limits 0 and 4 (the first four seconds), giving \(\dfrac{68}{3}\): M1 M0 A0. Condone no check that \(v > 0\) for \(3 \leqslant t \leqslant 4\) (it is).
1 mark
A1* \(\left(12 + 16 - \dfrac{64}{12}\right) - \left(9 + 9 - \dfrac{27}{12}\right) = \dfrac{272}{12} - \dfrac{189}{12} = \dfrac{83}{12}\)
cso. Both values, or an equivalent intermediate line, must be seen.

Question 2

6 marks

A remote-controlled car moves along a straight horizontal track. At time \(t = 0\) the car is at the point \(O\) on the track. At time \(t\) seconds, the velocity of the car is \(v\ \text{m s}^{-1}\), where
\[v = 6\sqrt{t} - 2t, \quad t \geqslant 0\]
Find the time at which the car returns to \(O\), and its speed at that instant.

Model answer

\(x = \displaystyle\int (6t^{\frac{1}{2}} - 2t)\,dt = 4t^{\frac{3}{2}} - t^2 + c\)
\(t = 0,\ x = 0\): \(c = 0\)
At \(O\): \(4t^{\frac{3}{2}} - t^2 = 0\), so \(t^{\frac{3}{2}}(4 - t^{\frac{1}{2}}) = 0\)
\(t^{\frac{1}{2}} = 4\), so \(t = 16\)
\(v = 6\sqrt{16} - 2 \times 16 = 24 - 32 = -8\)
The car returns to \(O\) when \(t = 16\), with speed \(8\ \text{m s}^{-1}\).

Explanation

Integrate the velocity to find the displacement, using \(x = 0\) when \(t = 0\), and then solve \(x = 0\) for the non-zero root by factorising out \(t^{\frac{3}{2}}\). Substituting \(t = 16\) into \(v\) gives \(-8\); the car is moving back towards \(O\), so its speed is 8. Common errors are giving a negative speed and solving \(v = 0\), which gives the time when the car turns round, \(t = 9\).

Mark scheme

1 mark
M1 Integrates \(v\), with at least one power increased by 1 (e.g. \(t^{\frac{1}{2}} \to t^{\frac{3}{2}}\)); or correct final answers of \(t = 16\) and a speed of 8 \(\text{m s}^{-1}\) with no working
Answers of \(t = 16\) and a speed of 8 \(\text{m s}^{-1}\) with no working score 2 marks (the A1 for 16 and the A1 for 8). M0 for differentiating.
1 mark
A1 \(x = 4t^{\frac{3}{2}} - t^2\) (the constant is 0 because \(x = 0\) when \(t = 0\))
Accept \(4t\sqrt{t} - t^2\).
1 mark
M1 (dM1) Sets their \(x = 0\) and solves for \(t ≠ 0\), e.g. \(t^{\frac{3}{2}}(4 - t^{\frac{1}{2}}) = 0\) or \(4\sqrt{t} = t\)
Depends on the first M mark (not on the A mark just before it).
1 mark
A1 \(t = 16\)
\(t = 0\) must not be given as a final answer (ignore it if it is rejected).
1 mark
M1 Substitutes their \(t\) into \(v\)
Their non-zero value of \(t\).
1 mark
A1 Speed \(= 8\ \text{m s}^{-1}\)
The speed must be positive: \(v = 24 - 32 = -8\) followed by a final answer of \(-8\) for the speed is A0. Accept "velocity \(-8\), so speed 8".

Question 3

7 marks

A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds, the acceleration of \(P\) is \((6t - 14)\ \text{m s}^{-2}\) in the positive \(x\)-direction. The particle passes through the origin \(O\) when \(t = 1\) and again when \(t = 5\).

Find an expression, in terms of \(t\), for the displacement of \(P\) from \(O\) at time \(t\) seconds.

Hence show that \(P\) is instantaneously at rest when \(t = 1\).

Model answer

\(v = \displaystyle\int (6t - 14)\,dt = 3t^2 - 14t + c\)
\(x = \displaystyle\int (3t^2 - 14t + c)\,dt = t^3 - 7t^2 + ct + d\)
\(t = 1,\ x = 0\): \(1 - 7 + c + d = 0\), so \(c + d = 6\)
\(t = 5,\ x = 0\): \(125 - 175 + 5c + d = 0\), so \(5c + d = 50\)
Subtracting: \(4c = 44\), so \(c = 11\) and \(d = -5\)
\(x = t^3 - 7t^2 + 11t - 5\)
When \(t = 1\): \(v = 3 - 14 + 11 = 0\), so \(P\) is instantaneously at rest when \(t = 1\).

Explanation

Integrate twice, keeping both constants of integration, then use the two times at which \(P\) is at \(O\) to form simultaneous equations for them. With \(c = 11\), the velocity at \(t = 1\) is zero, so \(P\) is at rest at the moment it passes through \(O\). A common error is to assume the velocity or displacement is zero when \(t = 0\), which the question does not say.

Mark scheme

1 mark
M1 Integrates the acceleration, with at least one power increased by 1. An answer with no working scores 0 marks
An answer with no working scores 0 marks: the constants must be found from the given conditions, and the last step is a show that. Use of the constant acceleration formulae (suvat) scores no marks in this question, because the acceleration is not constant.
1 mark
A1 \(v = 3t^2 - 14t + c\)
The constant must be present (it is not known yet).
1 mark
M1 Integrates their \(v\) to obtain a displacement with a second constant: \(x = t^3 - 7t^2 + ct + d\)
At least one power increased by 1. M0 if either constant is omitted.
1 mark
M1 (dM1) Uses \(x = 0\) at \(t = 1\) and at \(t = 5\) to form two equations, \(c + d = 6\) and \(5c + d = 50\), and solves them
Depends on the previous M mark.
1 mark
A1 \(x = t^3 - 7t^2 + 11t - 5\)
\(c = 11\), \(d = -5\). Accept \((t - 1)^2(t - 5)\).
1 mark
M1 Finds the velocity at \(t = 1\) using their \(c\): \(v = 3(1)^2 - 14(1) + 11\), or differentiates their \(x\) and substitutes \(t = 1\)
Their value of \(c\). Also accept: showing that \(x = (t - 1)^2(t - 5)\) has a repeated root at \(t = 1\), so \(\dfrac{dx}{dt} = 0\) there, with the derivative evaluated.
1 mark
A1* \(v = 3 - 14 + 11 = 0\), so \(P\) is instantaneously at rest when \(t = 1\)
cso, with a conclusion.

Question 4

8 marks

A particle \(P\) moves along a straight line. At time \(t\) seconds, the velocity of \(P\) is \(v\ \text{m s}^{-1}\), where
\[v = 3t^2 - 30t + 63, \quad t \geqslant 0\]
Find the total distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 8\).

Model answer

\(v = 0\): \(3t^2 - 30t + 63 = 0\), so \(t^2 - 10t + 21 = 0\), \((t - 3)(t - 7) = 0\)
\(P\) changes direction when \(t = 3\) and \(t = 7\).
\(x = \displaystyle\int v\,dt = t^3 - 15t^2 + 63t\) (taking \(x = 0\) when \(t = 0\))
\(x(3) = 27 - 135 + 189 = 81\)
\(x(7) = 343 - 735 + 441 = 49\)
\(x(8) = 512 - 960 + 504 = 56\)
Distances: \(0 \leqslant t \leqslant 3\): 81 m; \(3 \leqslant t \leqslant 7\): \(81 - 49 = 32\) m; \(7 \leqslant t \leqslant 8\): \(56 - 49 = 7\) m
Total distance \(= 81 + 32 + 7 = 120\) m

Explanation

The particle changes direction where \(v = 0\), at \(t = 3\) and \(t = 7\), so the integral must be split at these times and the magnitudes of the three displacements added. The marks are for finding the times, integrating, evaluating each interval and adding the distances. The common error is to integrate from 0 to 8 in one go, which gives the displacement, 56 m, not the distance.

Mark scheme

1 mark
M1 Sets \(v = 0\) and solves the three-term quadratic to find when \(P\) changes direction; or a correct final answer of 120 m with no working
An answer of 120 m with no working scores 1 mark (the final A1 only). A common error is to integrate from 0 to 8 without considering the changes of direction, giving 56 m: this scores at most the M1 A1 for the integration.
1 mark
A1 \(t = 3\) and \(t = 7\)
Both needed.
1 mark
M1 Integrates \(v\), with at least one power increased by 1
Independent of the first two marks. M0 for differentiating, or for using suvat.
1 mark
A1 \(x = t^3 - 15t^2 + 63t \; (+ c)\)
Taking \(x = 0\) at \(t = 0\) (the constant cancels in definite integrals).
1 mark
M1 (dM1) Finds the displacements over the separate intervals \(0 \leqslant t \leqslant 3\), \(3 \leqslant t \leqslant 7\) and \(7 \leqslant t \leqslant 8\), using their times (definite integrals or values of their \(x\) at \(t = 0, 3, 7, 8\))
Depends on the second M mark (the integration). Values of \(x\): \(x(3) = 81\), \(x(7) = 49\), \(x(8) = 56\).
1 mark
A1 Distances 81, 32 and 7 (or displacements 81, \(-32\) and 7)
All three needed. Accept the \(x\) values 81, 49 and 56 with the correct differences.
1 mark
M1 (ddM1) Adds the magnitudes of the three distances
Depends on the first and third M marks. Adding signed values (\(81 - 32 + 7\)) is M0.
1 mark
A1 Total distance \(= 120\) m
cao. Ignore units.

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