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

Practise 17 exam-style questions on Constant 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

3 marks

A particle \(P\) moves with constant acceleration \((1.6\mathbf{i} - 2.4\mathbf{j})\ \text{m s}^{-2}\). At time \(t = 0\), \(P\) has velocity \((-4\mathbf{i} + 6\mathbf{j})\ \text{m s}^{-1}\).

Show that \(P\) is instantaneously at rest at one instant, and find the value of \(t\) at that instant.

Model answer

\(\mathbf{v} = (-4\mathbf{i} + 6\mathbf{j}) + t(1.6\mathbf{i} - 2.4\mathbf{j}) = (-4 + 1.6t)\mathbf{i} + (6 - 2.4t)\mathbf{j}\)
\(\mathbf{i}\): \(-4 + 1.6t = 0\), so \(t = 2.5\)
\(\mathbf{j}\): \(6 - 2.4t = 0\), so \(t = 2.5\)
Both components are zero at the same instant, \(t = 2.5\), so \(\mathbf{v} = \mathbf{0}\): \(P\) is instantaneously at rest when \(t = 2.5\).

Explanation

A particle is at rest only when its whole velocity vector is zero, so both components must vanish at the same time. Here they do at \(t = 2.5\), because the acceleration is a negative multiple of the initial velocity. Checking only one component is the common error: it shows only that the motion is parallel to one of the unit vectors.

Mark scheme

1 mark
M1 Velocity at time \(t\): \(\mathbf{v} = (-4 + 1.6t)\mathbf{i} + (6 - 2.4t)\mathbf{j}\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the show-that needs the velocity at time \(t\) and both components. Column vectors are accepted throughout the working.
1 mark
M1 (dM1) Sets both components equal to zero and solves each: \(-4 + 1.6t = 0\) and \(6 - 2.4t = 0\)
Both components must be considered. Alternatively, shows \(\mathbf{v} = \mathbf{0}\) by substituting \(t = 2.5\) into both components.
1 mark
A1* Both components are zero when \(t = 2.5\) (and at no other time), so \(P\) is instantaneously at rest when \(t = 2.5\)
cso. A conclusion is needed. A0 if only one component is checked.

Question 2

5 marks

[In this question, \(\mathbf{i}\) is a horizontal unit vector due east and \(\mathbf{j}\) is a horizontal unit vector due north.]

A model aircraft flies at a constant height with constant acceleration. At time \(t = 0\), its velocity is \((3\mathbf{i} + 5\mathbf{j})\ \text{m s}^{-1}\). At time \(t = 6\) seconds, its velocity is \((-6\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-1}\).

Find the acceleration of the aircraft.

Show that, when \(t = 10\), the aircraft is moving due west, and find its speed at this time.

Model answer

\(\mathbf{a} = \dfrac{\mathbf{v} - \mathbf{u}}{t} = \dfrac{(-6\mathbf{i} + 2\mathbf{j}) - (3\mathbf{i} + 5\mathbf{j})}{6} = \dfrac{-9\mathbf{i} - 3\mathbf{j}}{6} = (-1.5\mathbf{i} - 0.5\mathbf{j})\ \text{m s}^{-2}\)
At \(t = 10\): \(\mathbf{v} = (3\mathbf{i} + 5\mathbf{j}) + 10(-1.5\mathbf{i} - 0.5\mathbf{j}) = (3 - 15)\mathbf{i} + (5 - 5)\mathbf{j} = -12\mathbf{i}\)
The velocity has no \(\mathbf{j}\)-component and a negative \(\mathbf{i}\)-component, so the aircraft is moving due west.
Speed \(= 12\ \text{m s}^{-1}\)

Explanation

The constant acceleration is the change in velocity divided by the time taken, worked out component by component. Due west means the velocity is a negative multiple of \(\mathbf{i}\), with no \(\mathbf{j}\)-component, which is what \(-12\mathbf{i}\) shows. A common error is to subtract the velocities the wrong way round.

Mark scheme

1 mark
M1 Uses \(\mathbf{a} = \dfrac{\mathbf{v} - \mathbf{u}}{t}\): \(\dfrac{(-6\mathbf{i} + 2\mathbf{j}) - (3\mathbf{i} + 5\mathbf{j})}{6}\); a correct acceleration of \((-1.5\mathbf{i} - 0.5\mathbf{j})\ \text{m s}^{-2}\) with no working scores the M1 and A1, but the show-that needs working
An answer of \((-1.5\mathbf{i} - 0.5\mathbf{j})\ \text{m s}^{-2}\) for the acceleration and \(12\ \text{m s}^{-1}\) for the speed with no working scores 3 marks (M1 A1 and the final B1); the show-that mark needs the velocity at \(t = 10\). M0 if \(\mathbf{u}\) and \(\mathbf{v}\) are subtracted the wrong way round and not corrected. Column vectors are accepted throughout the working.
1 mark
A1 \((-1.5\mathbf{i} - 0.5\mathbf{j})\ \text{m s}^{-2}\)
Accept \(\left(-\tfrac{3}{2}\mathbf{i} - \tfrac{1}{2}\mathbf{j}\right)\).
1 mark
M1 Velocity at \(t = 10\) using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with their \(\mathbf{a}\): \((3\mathbf{i} + 5\mathbf{j}) + 10(-1.5\mathbf{i} - 0.5\mathbf{j})\)
Alternatively from \(t = 6\): \((-6\mathbf{i} + 2\mathbf{j}) + 4\mathbf{a}\).
1 mark
A1* \(\mathbf{v} = -12\mathbf{i}\) (the \(\mathbf{j}\)-component is 0 and the \(\mathbf{i}\)-component is negative), so the aircraft is moving due west
cso. Both facts are needed: zero \(\mathbf{j}\)-component and negative \(\mathbf{i}\)-component, with a conclusion.
1 mark
B1 Speed \(12\ \text{m s}^{-1}\)
ft the magnitude of their velocity at \(t = 10\). Must be positive.

Question 3

6 marks

A particle \(P\) moves with constant acceleration \((2\mathbf{i} + \mathbf{j})\ \text{m s}^{-2}\). At time \(t = 0\), \(P\) has velocity \((-2\mathbf{i} - 6\mathbf{j})\ \text{m s}^{-1}\).

Find the values of \(t\) at which the speed of \(P\) is \(5\ \text{m s}^{-1}\).

Model answer

\(\mathbf{v} = (-2\mathbf{i} - 6\mathbf{j}) + t(2\mathbf{i} + \mathbf{j}) = (2t - 2)\mathbf{i} + (t - 6)\mathbf{j}\)
Speed 5: \((2t - 2)^2 + (t - 6)^2 = 25\)
\(4t^2 - 8t + 4 + t^2 - 12t + 36 = 25\)
\(5t^2 - 20t + 15 = 0\), so \(t^2 - 4t + 3 = 0\)
\((t - 1)(t - 3) = 0\)
\(t = 1\) or \(t = 3\)

Explanation

The speed is the magnitude of the velocity, so square both components, add them and set the total equal to \(5^2\). The result is a quadratic in \(t\) with two positive roots: the speed falls to 5 at \(t = 1\), drops below it, and rises back to 5 at \(t = 3\). A common error is to add the components without squaring.

Mark scheme

1 mark
M1 Velocity at time \(t\): \(\mathbf{v} = (-2 + 2t)\mathbf{i} + (t - 6)\mathbf{j}\); correct final answers of \(t = 1\) and \(t = 3\) with no working score 2 marks
Answers of \(t = 1\) and \(t = 3\) with no working score 2 marks (the two final A marks). Column vectors are accepted throughout the working.
1 mark
M1 Uses Pythagoras on their \(\mathbf{v}\) and sets the speed (or its square) equal to 5 (or 25): \((2t - 2)^2 + (t - 6)^2 = 25\)
M0 for \((2t - 2) + (t - 6) = 5\), or for subtracting the squares.
1 mark
A1 \(5t^2 - 20t + 15 = 0\), i.e. \(t^2 - 4t + 3 = 0\)
Any correct three-term quadratic.
1 mark
M1 (dM1) Solves their three-term quadratic
Depends on the previous M mark (not on the A mark just before it). Factorising: \((t - 1)(t - 3) = 0\).
1 mark
A1 \(t = 1\)
cao
1 mark
A1 \(t = 3\)
Depends on the third M mark. cao

Question 4

8 marks

A particle \(P\) moves with constant acceleration. At time \(t = 0\), \(P\) passes through the fixed origin \(O\). At time \(t = 2\) seconds, \(P\) is at the point with position vector \((4\mathbf{i} + 10\mathbf{j})\) m, and at time \(t = 6\) seconds, \(P\) is at the point with position vector \((24\mathbf{i} + 6\mathbf{j})\) m.

Find, in terms of \(\mathbf{i}\) and \(\mathbf{j}\), the velocity of \(P\) at time \(t = 0\) and the acceleration of \(P\).

Find the value of \(t\) when \(P\) is moving parallel to \(\mathbf{i}\), and the velocity of \(P\) at this time.

Model answer

\(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2\)
\(t = 2\): \(4\mathbf{i} + 10\mathbf{j} = 2\mathbf{u} + 2\mathbf{a}\), so \(\mathbf{u} + \mathbf{a} = 2\mathbf{i} + 5\mathbf{j}\)
\(t = 6\): \(24\mathbf{i} + 6\mathbf{j} = 6\mathbf{u} + 18\mathbf{a}\), so \(\mathbf{u} + 3\mathbf{a} = 4\mathbf{i} + \mathbf{j}\)
Subtracting: \(2\mathbf{a} = 2\mathbf{i} - 4\mathbf{j}\), so \(\mathbf{a} = (\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-2}\)
\(\mathbf{u} = (2\mathbf{i} + 5\mathbf{j}) - (\mathbf{i} - 2\mathbf{j}) = (\mathbf{i} + 7\mathbf{j})\ \text{m s}^{-1}\)
\(\mathbf{v} = (\mathbf{i} + 7\mathbf{j}) + t(\mathbf{i} - 2\mathbf{j}) = (1 + t)\mathbf{i} + (7 - 2t)\mathbf{j}\)
Parallel to \(\mathbf{i}\) when \(7 - 2t = 0\), so \(t = 3.5\), and then \(\mathbf{v} = 4.5\mathbf{i}\ \text{m s}^{-1}\).

Explanation

Each position gives a vector equation in the unknown vectors \(\mathbf{u}\) and \(\mathbf{a}\), and these are solved simultaneously just like scalar equations. A velocity parallel to \(\mathbf{i}\) has no \(\mathbf{j}\)-component, so the \(\mathbf{j}\)-component of \(\mathbf{v}\) is set to zero. A common error is to leave out the \(\tfrac{1}{2}\) in \(\tfrac{1}{2}\mathbf{a}t^2\).

Mark scheme

1 mark
M1 Uses \(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2\) at \(t = 2\): \(4\mathbf{i} + 10\mathbf{j} = 2\mathbf{u} + 2\mathbf{a}\); correct final answers of \(\mathbf{u} = (\mathbf{i} + 7\mathbf{j})\), \(\mathbf{a} = (\mathbf{i} - 2\mathbf{j})\), \(t = 3.5\) and \(4.5\mathbf{i}\ \text{m s}^{-1}\) with no working score 4 marks
Answers of \(\mathbf{u} = (\mathbf{i} + 7\mathbf{j})\ \text{m s}^{-1}\), \(\mathbf{a} = (\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-2}\), \(t = 3.5\) and \(\mathbf{v} = 4.5\mathbf{i}\ \text{m s}^{-1}\) with no working score 4 marks (the four A marks): the methods must be seen for the others. Column vectors are accepted throughout the working.
1 mark
M1 Uses \(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2\) at \(t = 6\): \(24\mathbf{i} + 6\mathbf{j} = 6\mathbf{u} + 18\mathbf{a}\)
M0 if \(\tfrac{1}{2}\) is missing.
1 mark
M1 (ddM1) Solves the two vector equations simultaneously (or the four component equations)
Depends on both previous M marks.
1 mark
A1 \(\mathbf{a} = (\mathbf{i} - 2\mathbf{j})\ \text{m s}^{-2}\)
Final answer in \(\mathbf{i}\), \(\mathbf{j}\) form.
1 mark
A1 \(\mathbf{u} = (\mathbf{i} + 7\mathbf{j})\ \text{m s}^{-1}\)
Depends on the third M mark. Final answer in \(\mathbf{i}\), \(\mathbf{j}\) form.
1 mark
M1 Velocity at time \(t\) with their \(\mathbf{u}\) and \(\mathbf{a}\), and sets the \(\mathbf{j}\)-component equal to zero: \(7 - 2t = 0\)
M0 for setting the \(\mathbf{i}\)-component to zero.
1 mark
A1 \(t = 3.5\)
cao
1 mark
A1 \(\mathbf{v} = 4.5\mathbf{i}\ \text{m s}^{-1}\)
Depends on the fourth M mark. A speed of 4.5 without the direction is A0.

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