| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a standard M1 SUVAT question with multiple parts requiring algebraic manipulation and graph sketching. Part (a) involves simple algebra with uniform acceleration, part (b) is routine graph sketching, part (c) uses area under velocity-time graph (trapezium area formula), and part (d) requires basic interpretation of inverse proportionality. All techniques are standard M1 fare with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) e.g. since acc" and decel" are uniform, time for decel" \(= \frac{1}{12.5}\) time for acc" | M1 | |
| \(\therefore\) decel" \(= 4\) seconds, so total time \(= 6 + 50 + 4 = 60\) seconds | M1 A1 | |
| (b) [velocity-time graph with trapezoid shape: acceleration from 0 to 6s, constant velocity from 6 to 56s, deceleration from 56 to 60s] | B3 | |
| (c) area under graph \(= \frac{1}{2}(6)(V) + 50V + \frac{1}{2}(4)(V) = 1320\) | M1 | |
| \(55V = 1320\) so \(V = 24 \text{ ms}^{-1}\) | M1 A1 | |
| (d) car accelerates more quickly at first, but acceleration decreases throughout the six seconds | B2 | (11) |
**(a)** e.g. since acc" and decel" are uniform, time for decel" $= \frac{1}{12.5}$ time for acc" | M1 |
$\therefore$ decel" $= 4$ seconds, so total time $= 6 + 50 + 4 = 60$ seconds | M1 A1 |
**(b)** [velocity-time graph with trapezoid shape: acceleration from 0 to 6s, constant velocity from 6 to 56s, deceleration from 56 to 60s] | B3 |
**(c)** area under graph $= \frac{1}{2}(6)(V) + 50V + \frac{1}{2}(4)(V) = 1320$ | M1 |
$55V = 1320$ so $V = 24 \text{ ms}^{-1}$ | M1 A1 |
**(d)** car accelerates more quickly at first, but acceleration decreases throughout the six seconds | B2 | (11)5. A car on a straight test track starts from rest and accelerates to a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 6 seconds. The car maintains this speed for a further 50 seconds before decelerating to rest.
In a simple model of this motion, the acceleration and deceleration are assumed to be uniform and the magnitude of the deceleration to be 1.5 times that of the acceleration.
\begin{enumerate}[label=(\alph*)]
\item Show that the total time for which the car is moving is 60 seconds.
\item Sketch a velocity-time graph for this journey.
Given that the total distance travelled is 1320 metres,
\item find $V$.
In a more sophisticated model, the acceleration is assumed to be inversely proportional to the velocity of the car.
\item Explain how the acceleration would vary during the first six seconds under this model.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q5 [11]}}