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

Practise 18 exam-style questions on Constant 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.

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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 skateboarder moves in a straight line down a gentle slope. She passes a post \(P\) with speed \(4\ \text{m s}^{-1}\) and moves with constant acceleration \(0.5\ \text{m s}^{-2}\). She passes a second post \(Q\), which is 60 m from \(P\).

Find the time she takes to travel from \(P\) to \(Q\), giving your answer to 3 significant figures.

[Constant acceleration: \(v = u + at\), \(s = ut + \tfrac{1}{2}at^2\), \(s = vt - \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\), \(s = \tfrac{1}{2}(u + v)t\)]

Model answer

\(u = 4\), \(a = 0.5\), \(s = 60\)
\(s = ut + \tfrac{1}{2}at^2\): \(60 = 4t + 0.25t^2\)
\(t^2 + 16t - 240 = 0\)
\(t = \dfrac{-16 + \sqrt{16^2 + 4 \times 240}}{2} = \dfrac{-16 + \sqrt{1216}}{2} = 9.4356...\)
The time is 9.44 s (3 s.f.).

Explanation

With \(u\), \(a\) and \(s\) known and \(t\) wanted, \(s = ut + \tfrac{1}{2}at^2\) gives a quadratic in \(t\), and only the positive root makes sense. Finding the speed at \(Q\) first with \(v^2 = u^2 + 2as\), then using \(v = u + at\), is an equally good route. A common error is to forget the factor \(\tfrac{1}{2}\) in \(\tfrac{1}{2}at^2\).

Mark scheme

1 mark
M1 Uses \(s = ut + \tfrac{1}{2}at^2\): \(60 = 4t + \tfrac{1}{2}(0.5)t^2\); or a correct final answer of 9.44 s with no working
An answer of 9.44 s with no working scores 3 marks (accept awrt 9.44, i.e. [9.435, 9.445)). Alternative method: \(v^2 = 4^2 + 2(0.5)(60) = 76\), then \(t = \dfrac{\sqrt{76} - 4}{0.5}\) scores M1 M1 for the two steps.
1 mark
M1 (dM1) Solves the three-term quadratic \(t^2 + 16t - 240 = 0\) (or their equivalent) by the formula or completing the square
Depends on the first M mark. A calculator solution of a correct quadratic is acceptable.
1 mark
A1 awrt 9.44 s, i.e. [9.435, 9.445)
Exact value \(-8 + 4\sqrt{19}\). The negative root must be rejected (or ignored); A0 if both roots are given as answers.

Question 2

5 marks

A particle moves in a straight line with constant acceleration \(a\ \text{m s}^{-2}\). In a time of \(t\) seconds its velocity increases from \(u\ \text{m s}^{-1}\) to \(v\ \text{m s}^{-1}\) and its displacement is \(s\) metres. Figure 1 shows the velocity-time graph for this motion.

By considering the graph, show that \(v = u + at\) and that \(s = \tfrac{1}{2}(u + v)t\).

Hence show that \(v^2 = u^2 + 2as\).

A sketch of a velocity-time graph, with velocity (m s⁻¹) on the vertical axis and time (s) on the horizontal axis. A straight line rises from u on the velocity axis to v at time t. The region under the line, between the axes and the dashed vertical line at time t, is shaded. A dashed line joins v on the velocity axis to the end of the line.

Model answer

The gradient of a velocity-time graph is the acceleration:
\(a = \dfrac{v - u}{t}\), so \(at = v - u\) and \(v = u + at\)
The area under the graph is the displacement. The area is a trapezium with parallel sides \(u\) and \(v\) and width \(t\):
\(s = \tfrac{1}{2}(u + v)t\)
From the first result, \(t = \dfrac{v - u}{a}\). Substituting:
\(s = \tfrac{1}{2}(u + v)\dfrac{(v - u)}{a}\)
\(2as = (v + u)(v - u) = v^2 - u^2\)
\(v^2 = u^2 + 2as\)

Explanation

On a velocity-time graph the gradient is the acceleration and the area is the displacement, which gives the first two constant-acceleration formulae directly. Eliminating \(t\) between them, and using the difference of two squares, gives the formula without \(t\). The printed results must be reached with every step shown, not quoted.

Mark scheme

1 mark
B1* Gradient of the graph = acceleration: \(a = \dfrac{v - u}{t}\), so \(v = u + at\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: these are show-that results and every step must be seen. Must refer to the gradient (or the rate of change of velocity) explicitly.
1 mark
M1 Displacement = area under the graph: a trapezium with parallel sides \(u\) and \(v\) and width \(t\) (or a rectangle \(ut\) plus a triangle \(\tfrac{1}{2}(v - u)t\))
The area must be identified as the displacement.
1 mark
A1* \(s = \tfrac{1}{2}(u + v)t\)
cso. From the rectangle and triangle: \(ut + \tfrac{1}{2}(v - u)t = \tfrac{1}{2}(u + v)t\) must be seen.
1 mark
M1 Eliminates \(t\) using \(t = \dfrac{v - u}{a}\) from the first result: \(s = \tfrac{1}{2}(u + v) \times \dfrac{v - u}{a}\)
Using the first two results only (not quoting \(v^2 = u^2 + 2as\) or another suvat equation).
1 mark
A1* \(2as = (v + u)(v - u) = v^2 - u^2\), so \(v^2 = u^2 + 2as\)
cso. At least one line showing the difference of two squares (or the expansion) is needed before the printed result.

Question 3

7 marks

A particle \(P\) moves along a straight line with constant acceleration. At time \(t = 0\), \(P\) passes through the point \(O\) on the line with velocity \(8\ \text{m s}^{-1}\) in the positive direction. The acceleration of \(P\) is \(2\ \text{m s}^{-2}\) in the negative direction.

Find the distance travelled by \(P\) in the first 6 seconds after it passes \(O\).

Find all the times at which \(P\) is 12 m from \(O\). Give any answer that is not exact to 3 significant figures.

[Constant acceleration: \(v = u + at\), \(s = ut + \tfrac{1}{2}at^2\), \(s = vt - \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\), \(s = \tfrac{1}{2}(u + v)t\)]

Model answer

\(s = 8t + \tfrac{1}{2}(-2)t^2 = 8t - t^2\), \(v = 8 - 2t\)
\(P\) is at rest when \(8 - 2t = 0\), i.e. \(t = 4\), so it turns back before \(t = 6\).
\(s(4) = 32 - 16 = 16\), \(s(6) = 48 - 36 = 12\)
Distance in the first 6 s \(= 16 + (16 - 12) = 20\) m
\(P\) is 12 m from \(O\) when \(s = 12\) or \(s = -12\).
\(8t - t^2 = 12\): \(t^2 - 8t + 12 = 0\), \((t - 2)(t - 6) = 0\), \(t = 2\) or \(t = 6\)
\(8t - t^2 = -12\): \(t^2 - 8t - 12 = 0\), \(t = \dfrac{8 \pm \sqrt{112}}{2} = 4 \pm 2\sqrt{7}\); \(t > 0\) so \(t = 4 + 2\sqrt{7} = 9.29\)
The times are \(t = 2\), \(t = 6\) and \(t = 9.29\) (3 s.f.).

Explanation

\(P\) stops at \(t = 4\) and comes back, so the distance travelled in 6 s is 16 m out plus 4 m back, not the 12 m displacement. Being 12 m from \(O\) means \(s = 12\) or \(s = -12\): the first gives the two times either side of the turning point, and the second gives a later time after \(P\) has passed back through \(O\). A common error is to miss the case \(s = -12\).

Mark scheme

1 mark
M1 Finds when \(P\) is at rest, \(0 = 8 - 2t\) so \(t = 4\), or otherwise recognises that \(P\) changes direction before \(t = 6\); correct final answers of 20 m and \(t = 2\), 6 and 9.29 with no working score 3 marks
Answers of 20 m and \(t = 2\), \(t = 6\) and \(t = 9.29\) with no working score 3 marks (the A1 for 20 m, the A1 for 2 and 6, and the final A1): the methods must be seen for the other marks.
1 mark
A1 Displacements \(s(4) = 16\) and \(s(6) = 12\) (both)
From \(s = 8t - t^2\).
1 mark
A1 Distance \(= 16 + (16 - 12) = 20\) m
Depends on the first M mark. A0 for 12 m, the displacement at \(t = 6\) (a common error).
1 mark
M1 Solves \(8t - t^2 = 12\), a three-term quadratic (\(t^2 - 8t + 12 = 0\))
By factorising, the formula or completing the square.
1 mark
A1 \(t = 2\) and \(t = 6\)
Both needed.
1 mark
M1 Considers \(P\) on the other side of \(O\): solves \(8t - t^2 = -12\) (\(t^2 - 8t - 12 = 0\))
Must use \(s = -12\). This mark is for realising that "12 m from \(O\)" includes the negative side.
1 mark
A1 \(t = 4 + 2\sqrt{7}\), awrt 9.29, i.e. [9.285, 9.295), only
The negative root \(4 - 2\sqrt{7}\) must be rejected or ignored. Accept \(4 + \sqrt{28}\).

Question 4

8 marks

A car is moving at a constant speed of \(U\ \text{m s}^{-1}\) along a straight horizontal road when the driver sees a hazard ahead. In a model,
• the car continues at \(U\ \text{m s}^{-1}\) for 0.75 seconds (the driver's reaction time),
• the driver then brakes and the car decelerates uniformly at \(6.4\ \text{m s}^{-2}\) until it stops.

Show that the total distance, \(d\) metres, travelled by the car from the moment the driver sees the hazard until it stops is given by \(d = 0.75U + \dfrac{U^2}{12.8}\).

The driver sees a hazard 50 m ahead. Find the greatest speed at which the car can be travelling if it is to stop within the 50 m. Give your answer in \(\text{km h}^{-1}\) to 3 significant figures.

State one way in which the model could be refined.

[Constant acceleration: \(v = u + at\), \(s = ut + \tfrac{1}{2}at^2\), \(s = vt - \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\), \(s = \tfrac{1}{2}(u + v)t\)]

Model answer

During the reaction time the car moves at constant speed: distance \(= 0.75U\)
Braking: \(v^2 = u^2 + 2as\), \(0 = U^2 - 2 \times 6.4 \times s\), so \(s = \dfrac{U^2}{12.8}\)
\(d = 0.75U + \dfrac{U^2}{12.8}\)
For the car to stop in 50 m: \(0.75U + \dfrac{U^2}{12.8} = 50\)
\(U^2 + 9.6U - 640 = 0\)
\(U = \dfrac{-9.6 + \sqrt{9.6^2 + 2560}}{2} = 20.9496...\)
\(20.9496... \times 3.6 = 75.4\ \text{km h}^{-1}\) (3 s.f.)
Refinement: model the deceleration as varying with the road surface, rather than constant.

Explanation

The stopping distance has two parts: the reaction distance at constant speed and the braking distance from \(v^2 = u^2 + 2as\) with \(v = 0\). Setting it equal to 50 m gives a quadratic in \(U\), whose positive root is converted from \(\text{m s}^{-1}\) to \(\text{km h}^{-1}\) by multiplying by 3.6. A refinement must change an assumption of this model, such as the constant deceleration or the fixed reaction time.

Mark scheme

1 mark
M1 Distance during the reaction time \(= 0.75U\) (constant speed). The formula for \(d\) must be shown: an answer with no working scores 0 marks for it, but a correct final answer of \(75.4\ \text{km h}^{-1}\) with no working scores 2 marks
An answer of \(75.4\ \text{km h}^{-1}\) with no working scores 2 marks (the A1 marks for \(U\) and for the conversion; accept awrt 75.4, i.e. [75.35, 75.45)); the show-that marks need every step.
1 mark
M1 Braking distance from \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = U\), \(a = -6.4\): \(0 = U^2 - 12.8s\)
Condone a sign error for the M mark.
1 mark
A1* \(d = 0.75U + \dfrac{U^2}{12.8}\)
Depends on both M marks. cso: the two distances must be added, with no errors seen. Accept the exact equivalent \(d = \dfrac{3U}{4} + \dfrac{5U^2}{64}\).
1 mark
M1 Sets \(0.75U + \dfrac{U^2}{12.8} = 50\) (or \(\leqslant 50\)) and forms a three-term quadratic: \(U^2 + 9.6U - 640 = 0\)
Using the printed formula is fine even if it was not shown. Accept any exact equivalent quadratic, e.g. \(5U^2 + 48U - 3200 = 0\).
1 mark
M1 (dM1) Solves their three-term quadratic
Depends on the previous M mark. Exact values are expected in the working.
1 mark
A1 \(U =\) awrt 20.9, i.e. [20.85, 20.95), or the exact value \(-4.8 + \sqrt{663.04}\)
The negative root must be rejected or ignored.
1 mark
A1 \(75.4\ \text{km h}^{-1}\) (awrt 75.4, i.e. [75.35, 75.45))
Depends on the second M mark (the solving). From \(U \times \dfrac{3600}{1000}\). No value of \(g\) is used, so a more accurate value that rounds to 75.4 (e.g. 75.42) is accepted. A0 for answers in \(\text{m s}^{-1}\) only.
1 mark
B1 A valid refinement, e.g. the reaction time varies between drivers; the deceleration may not be constant (it depends on the road surface, tyres or brakes); include the length of the car (do not model it as a particle)
B0 for "the road may not be horizontal" (not part of this model), for any mention of mass or weight, and for "use a more accurate value of g". If more than one refinement is given, all must be correct.

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