Variable acceleration with vectors - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 17 exam-style questions on Variable acceleration with vectors 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.

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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

4 marks

In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively.

A drone flies at a constant height above horizontal ground. At time \(t\) seconds, its velocity is \(\mathbf{v}\ \text{m s}^{-1}\), where
\[\mathbf{v} = (t^2 - 9t + 18)\mathbf{i} + (5 - t)\mathbf{j}, \quad 0 \leqslant t \leqslant 8\]
Find the value of \(t\) at which the drone is moving due north.

Model answer

Due north: the \(\mathbf{i}\)-component is zero and the \(\mathbf{j}\)-component is positive.
\(t^2 - 9t + 18 = 0\), so \((t - 3)(t - 6) = 0\), \(t = 3\) or \(t = 6\)
\(t = 3\): \(\mathbf{v} = 2\mathbf{j}\), due north
\(t = 6\): \(\mathbf{v} = -\mathbf{j}\), due south
So the drone moves due north when \(t = 3\).

Explanation

Moving due north needs the east component of the velocity to be zero and the north component to be positive. The quadratic gives two times, but at \(t = 6\) the north component is negative, so the drone is then moving due south. Common errors are setting the wrong component to zero and not checking the sign of the other component.

Mark scheme

1 mark
M1 Sets the \(\mathbf{i}\)-component of \(\mathbf{v}\) equal to zero and solves the three-term quadratic; or a correct final answer of \(t = 3\) with no working
An answer of \(t = 3\) with no working scores 2 marks (M1 A1): the rejection of \(t = 6\) needs to be seen for the last two marks. A common error is to set the \(\mathbf{j}\)-component equal to zero: this scores no marks.
1 mark
A1 \(t = 3\) and \(t = 6\)
From \((t - 3)(t - 6) = 0\). Both needed.
1 mark
M1 Considers the sign of the \(\mathbf{j}\)-component at their times: \(\mathbf{v}(3) = 2\mathbf{j}\), \(\mathbf{v}(6) = -\mathbf{j}\)
Independent of the previous marks.
1 mark
A1 \(t = 3\) only, because when \(t = 6\) the drone is moving due south
cso. A final answer of \(t = 3\) and \(t = 6\) is A0.

Question 2

6 marks

A particle \(P\) moves in a horizontal plane. At time \(t\) seconds, the acceleration of \(P\) is \(\mathbf{a}\ \text{m s}^{-2}\), where
\[\mathbf{a} = (4 - 2t)\mathbf{i} + 3\sqrt{t}\,\mathbf{j}, \quad t \geqslant 0\]
When \(t = 0\), the velocity of \(P\) is \((-3\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\).

Find the exact speed of \(P\) when \(t = 4\).

Model answer

\(\mathbf{v} = \displaystyle\int \left((4 - 2t)\mathbf{i} + 3t^{\frac{1}{2}}\mathbf{j}\right)dt = (4t - t^2)\mathbf{i} + 2t^{\frac{3}{2}}\mathbf{j} + \mathbf{c}\)
\(t = 0\): \(\mathbf{c} = -3\mathbf{i} + 2\mathbf{j}\)
\(\mathbf{v} = (4t - t^2 - 3)\mathbf{i} + (2t^{\frac{3}{2}} + 2)\mathbf{j}\)
When \(t = 4\): \(\mathbf{v} = (16 - 16 - 3)\mathbf{i} + (16 + 2)\mathbf{j} = -3\mathbf{i} + 18\mathbf{j}\)
Speed \(= \sqrt{(-3)^2 + 18^2} = \sqrt{333} = 3\sqrt{37}\ \text{m s}^{-1}\)

Explanation

Integrate each component of the acceleration, add the initial velocity as the constant of integration, then substitute \(t = 4\) and find the magnitude. The method marks are for integrating, using the initial velocity and using Pythagoras. Common errors are omitting the constant and giving a decimal when the exact speed is asked for.

Mark scheme

1 mark
M1 Integrates \(\mathbf{a}\), with at least one power increased by 1 in each component; or a correct final answer of \(3\sqrt{37}\ \text{m s}^{-1}\) with no working
An answer of \(3\sqrt{37}\) (or \(\sqrt{333}\)) \(\text{m s}^{-1}\) with no working scores 2 marks (the last two A marks). Integrating: at least one power increased by 1 in each component (M0 if they clearly just multiply by t, or use constant acceleration formulae). Accept column vectors throughout the working. M0 if the i and j are lost and not recovered.
1 mark
A1 \(\mathbf{v} = (4t - t^2)\mathbf{i} + 2t^{\frac{3}{2}}\mathbf{j} \; (+ \mathbf{c})\)
The constant may be missing for this mark.
1 mark
M1 (dM1) Uses \(\mathbf{v} = -3\mathbf{i} + 2\mathbf{j}\) when \(t = 0\) to find the constant and substitutes \(t = 4\)
Depends on the first M mark (not on the A mark just before it). M0 if the constant is omitted altogether.
1 mark
A1 \(\mathbf{v} = -3\mathbf{i} + 18\mathbf{j}\) when \(t = 4\)
From \(\mathbf{v} = (4t - t^2 - 3)\mathbf{i} + (2t^{\frac{3}{2}} + 2)\mathbf{j}\).
1 mark
M1 Uses Pythagoras, including the square root, on their velocity at \(t = 4\)
Independent of the previous marks, but must be applied to a velocity found by integration.
1 mark
A1 Speed \(= \sqrt{333} = 3\sqrt{37}\ \text{m s}^{-1}\)
Exact value required: the decimal 18.2 alone is A0. Accept \(\sqrt{333}\).

Question 3

7 marks

A particle \(P\) moves in a horizontal plane. At time \(t\) seconds, the velocity of \(P\) is \(\mathbf{v}\ \text{m s}^{-1}\), where
\[\mathbf{v} = (3t^2 + 2)\mathbf{i} + (8 - 2t)\mathbf{j}, \quad t \geqslant 0\]
When \(t = 0\), \(P\) is at the point with position vector \((-48\mathbf{i} - 9\mathbf{j})\) m relative to a fixed origin \(O\).

Find the distance of \(P\) from \(O\) at the instant when \(P\) is moving parallel to \(\mathbf{i}\).

Model answer

Parallel to \(\mathbf{i}\): \(8 - 2t = 0\), so \(t = 4\) (the \(\mathbf{i}\)-component \(3t^2 + 2\) is never zero)
\(\mathbf{r} = \displaystyle\int \mathbf{v}\,dt = (t^3 + 2t)\mathbf{i} + (8t - t^2)\mathbf{j} + \mathbf{c}\)
\(t = 0\): \(\mathbf{c} = -48\mathbf{i} - 9\mathbf{j}\)
\(\mathbf{r} = (t^3 + 2t - 48)\mathbf{i} + (8t - t^2 - 9)\mathbf{j}\)
When \(t = 4\): \(\mathbf{r} = (64 + 8 - 48)\mathbf{i} + (32 - 16 - 9)\mathbf{j} = 24\mathbf{i} + 7\mathbf{j}\)
Distance \(= \sqrt{24^2 + 7^2} = 25\) m

Explanation

Moving parallel to \(\mathbf{i}\) means the \(\mathbf{j}\)-component of the velocity is zero, which happens at \(t = 4\). Integrate the velocity, using the starting position as the constant, then find the magnitude of the position vector at \(t = 4\). Common errors are omitting the constant of integration and setting the wrong component to zero.

Mark scheme

1 mark
B1 \(P\) is moving parallel to \(\mathbf{i}\) when the \(\mathbf{j}\)-component of \(\mathbf{v}\) is zero: \(t = 4\). An answer of 25 m with no working scores 1 mark (the final A1 only)
An answer of 25 m with no working scores 1 mark (the final A1 only). Setting the \(\mathbf{i}\)-component to zero (no real roots) is B0.
1 mark
M1 Integrates \(\mathbf{v}\), with at least one power increased by 1 in each component
Integrating: at least one power increased by 1 in each component (M0 if they clearly just multiply by t, or use constant acceleration formulae). Accept column vectors throughout the working. M0 if the i and j are lost and not recovered.
1 mark
A1 \(\mathbf{r} = (t^3 + 2t)\mathbf{i} + (8t - t^2)\mathbf{j} \; (+ \mathbf{c})\)
1 mark
M1 (dM1) Uses \(\mathbf{r} = -48\mathbf{i} - 9\mathbf{j}\) when \(t = 0\) for the constant and substitutes their \(t\)
Depends on the M mark for integrating (not on the A mark just before it). M0 if the constant is omitted.
1 mark
A1 \(\mathbf{r} = 24\mathbf{i} + 7\mathbf{j}\)
1 mark
M1 Uses Pythagoras, including the square root, on their position vector
M0 if applied to the velocity.
1 mark
A1 Distance \(= 25\) m
cao.

Question 4

8 marks

A particle \(P\) moves in a horizontal plane. At time \(t\) seconds, the acceleration of \(P\) is \(\big((6 - 4t)\mathbf{i} - 3\mathbf{j}\big)\ \text{m s}^{-2}\). When \(t = 0\), the velocity of \(P\) is \((2\mathbf{i} + 9\mathbf{j})\ \text{m s}^{-1}\).

Find the value of \(t\) at which \(P\) is travelling in the direction of \((2\mathbf{i} + \mathbf{j})\).

Model answer

\(\mathbf{v} = \displaystyle\int \left((6 - 4t)\mathbf{i} - 3\mathbf{j}\right)dt = (6t - 2t^2)\mathbf{i} - 3t\mathbf{j} + \mathbf{c}\), with \(\mathbf{c} = 2\mathbf{i} + 9\mathbf{j}\)
\(\mathbf{v} = (2 + 6t - 2t^2)\mathbf{i} + (9 - 3t)\mathbf{j}\)
Parallel to \(2\mathbf{i} + \mathbf{j}\): \(2 + 6t - 2t^2 = 2(9 - 3t)\)
\(2t^2 - 12t + 16 = 0\), so \(t^2 - 6t + 8 = 0\), \((t - 2)(t - 4) = 0\)
\(t = 2\): \(\mathbf{v} = 6\mathbf{i} + 3\mathbf{j} = 3(2\mathbf{i} + \mathbf{j})\), in the direction of \(2\mathbf{i} + \mathbf{j}\)
\(t = 4\): \(\mathbf{v} = -6\mathbf{i} - 3\mathbf{j} = -3(2\mathbf{i} + \mathbf{j})\), in the opposite direction
So \(t = 2\).

Explanation

Integrate the acceleration and add the initial velocity, then use the fact that the \(\mathbf{i}\)-component must be twice the \(\mathbf{j}\)-component. Both roots of the quadratic make \(\mathbf{v}\) parallel to \(2\mathbf{i} + \mathbf{j}\), but only \(t = 2\) gives a positive multiple, so only then is \(P\) moving in that direction. Common errors are equating the components to 2 and 1 and not rejecting \(t = 4\).

Mark scheme

1 mark
M1 Integrates the acceleration, with at least one power increased by 1 in each component; or a correct final answer of \(t = 2\) with no working
An answer of \(t = 2\) with no working scores 2 marks (the last two A marks). Integrating: at least one power increased by 1 in each component (M0 if they clearly just multiply by t, or use constant acceleration formulae). Accept column vectors throughout the working. M0 if the i and j are lost and not recovered.
1 mark
A1 \(\mathbf{v} = (2 + 6t - 2t^2)\mathbf{i} + (9 - 3t)\mathbf{j}\)
The constant \(2\mathbf{i} + 9\mathbf{j}\) must be included.
1 mark
M1 Uses a correct direction condition on their \(\mathbf{v}\) to form an equation in \(t\) only: the \(\mathbf{i}\)-component is twice the \(\mathbf{j}\)-component, or \(\mathbf{v} = \lambda(2\mathbf{i} + \mathbf{j})\)
Independent of the A mark before it. Setting the components equal to 2 and 1, or equating \(\mathbf{v}\) to \(2\mathbf{i} + \mathbf{j}\), is M0. Condone the reciprocal ratio for the M mark.
1 mark
A1 \(2 + 6t - 2t^2 = 2(9 - 3t)\), i.e. \(t^2 - 6t + 8 = 0\) oe
1 mark
M1 (dM1) Solves their three-term quadratic
Depends on the previous M mark (not on the A mark just before it).
1 mark
A1 \(t = 2\) and \(t = 4\)
Both needed.
1 mark
M1 Checks the direction of \(\mathbf{v}\) at each of their times: \(\mathbf{v}(2) = 6\mathbf{i} + 3\mathbf{j} = 3(2\mathbf{i} + \mathbf{j})\) and \(\mathbf{v}(4) = -6\mathbf{i} - 3\mathbf{j} = -3(2\mathbf{i} + \mathbf{j})\)
Or finds the sign of \(\lambda\) at each time.
1 mark
A1 \(t = 2\) only: when \(t = 4\), \(P\) moves in the opposite direction, \(-(2\mathbf{i} + \mathbf{j})\)
cso. A final answer of \(t = 2\) and \(t = 4\) is A0.

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