| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Sketch velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question requiring application of SUVAT equations and velocity-time graph interpretation. Part (a) involves calculating two velocities and drawing three connected line segments. Part (b) requires finding areas under the graph (trapeziums/triangles). All steps are routine and mechanical with no problem-solving insight needed, making it easier than average but not trivial due to the multi-stage calculation. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 13(a) | Draws first stage of graph correctly. | AO3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Line moving down by 6 ms-1 in 5s | AO3.3 | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| end of the second stage. | AO2.2a | B1F |
| Answer | Marks |
|---|---|
| 13(b) | Finds at least one non-rectangular |
| Answer | Marks | Guidance |
|---|---|---|
| (backwards) | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| from areas above axis. OE | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| correctly. CAO. Units not required | AO3.2a | A1 |
| Total | 6 | |
| Q | Marking Instructions | AO |
Question 13:
--- 13(a) ---
13(a) | Draws first stage of graph correctly. | AO3.3 | B1 | Line connecting (0, 0) to (4, 3).
Line connecting (4, 3) to (9, -3).
Line connecting (9, -3) to (20, -3).
Draws second stage correctly
Line moving down by 6 ms-1 in 5s | AO3.3 | B1F
Deduces final stage correctly
Horizontal line of length 33/v where
v is the speed of the vehicle at the
end of the second stage. | AO2.2a | B1F
--- 13(b) ---
13(b) | Finds at least one non-rectangular
area.
As typical solution
Or t between 0 & 6.5: Area of
triangle
= 3 x 6.5 ÷ 2 = 9.75 m (forward)
t between 6.5 & 20: Area of
trapezium
= 0.5(13.5 + 11) x (-3) = -36.75 m
(backwards) | AO1.1a | M1 | t between 0 & 4: Area of triangle
= 3 x 4 ÷ 2 = 6 m (forward)
t between 4 & 6.5: Area of triangle
= 3 x 2.5 ÷ 2 = 3.75 m (forward)
t between 6.5 & 9: Area of triangle
= 2.5 x -3 ÷ 2 = -3.75 m
(backwards)
t between 9 and 20: Area of rect.
= -3 x 11 = -33 m (backwards)
Displacement from O is
9.75 – 36.75 = –27 m
Distance is 27m
Subtracts their areas below axis
from areas above axis. OE | AO1.1a | M1
Calculates distance from P
correctly. CAO. Units not required | AO3.2a | A1
Total | 6
Q | Marking Instructions | AO | Marks | Typical SolutionA vehicle, which begins at rest at point $P$, is travelling in a straight line.
For the first $4$ seconds the vehicle moves with a constant acceleration of $0.75\,\mathrm{m}\,\mathrm{s}^{-2}$
For the next $5$ seconds the vehicle moves with a constant acceleration of $-1.2\,\mathrm{m}\,\mathrm{s}^{-2}$
The vehicle then immediately stops accelerating, and travels a further $33\,\mathrm{m}$ at constant speed.
\begin{enumerate}[label=(\alph*)]
\item Draw a velocity-time graph for this journey on the grid below.
[3 marks]
\includegraphics{figure_13}
\item Find the distance of the car from $P$ after $20$ seconds.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q13 [6]}}