| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 problem requiring a speed-time graph sketch and solving simultaneous equations from area (distance) and time constraints. While it involves multiple stages and algebraic manipulation with the 2:1 deceleration-to-acceleration ratio, it follows a well-practiced template with no novel insight required—slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Trapezium shape (velocity-time graph) | B1 | Shape |
| B1 | Relative gradient — RHS steeper than LHS | |
| \(17\) and \(170\) shown on axes | B1 | 17 and 170 shown |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(T; 2T\) | B1 | Correct ratios of times for acceleration and deceleration seen or implied |
| \(\dfrac{170+(170-3T)}{2} \cdot 17 = 2125\) or \(\frac{1}{2}\times T_1\times 17 + 17(170-(T_1+T_2))+\frac{1}{2}\times 17\times T_2 = 2125\) or \(2125 = \dfrac{17}{2}(170+T')\) | M1 | Form an equation for total distance with their times |
| A2 | \(-1\) each error | |
| \(T = 30\) or \(T_1 + T_2 = 90\) | A1 | Use their equation and the correct ratio to find the value for time decelerating or total time accelerating and decelerating |
| M1 | Use of \(v = u + at\) or equivalent | |
| decel \(= \dfrac{17}{30}\) oe | A1 | \((0.5\dot{6})\) 3sf or better. Must be positive |
| Total: (7) |
# Question 5(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Trapezium shape (velocity-time graph) | B1 | Shape |
| | B1 | Relative gradient — RHS steeper than LHS |
| $17$ and $170$ shown on axes | B1 | 17 and 170 shown |
| **Total: (3)** | | |
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# Question 5(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $T; 2T$ | B1 | Correct ratios of times for acceleration and deceleration seen or implied |
| $\dfrac{170+(170-3T)}{2} \cdot 17 = 2125$ or $\frac{1}{2}\times T_1\times 17 + 17(170-(T_1+T_2))+\frac{1}{2}\times 17\times T_2 = 2125$ or $2125 = \dfrac{17}{2}(170+T')$ | M1 | Form an equation for total distance with their times |
| | A2 | $-1$ each error |
| $T = 30$ or $T_1 + T_2 = 90$ | A1 | Use their equation and the correct ratio to find the value for time decelerating or total time accelerating and decelerating |
| | M1 | Use of $v = u + at$ or equivalent |
| decel $= \dfrac{17}{30}$ oe | A1 | $(0.5\dot{6})$ 3sf or better. Must be positive |
| **Total: (7)** | | |
---5. A car travelling along a straight horizontal road takes 170 s to travel between two sets of traffic lights at $A$ and $B$ which are 2125 m apart. The car starts from rest at $A$ and moves with constant acceleration until it reaches a speed of $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then maintains this speed before moving with constant deceleration, coming to rest at $B$. The magnitude of the deceleration is twice the magnitude of the acceleration.
\begin{enumerate}[label=(\alph*)]
\item Sketch, in the space below, a speed-time graph for the motion of the car between $A$ and $B$.
\item Find the deceleration of the car.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q5 [10]}}