| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| 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 problem requiring systematic application of kinematic equations across three motion phases. While it involves multiple stages and an unknown V to solve, the approach is methodical: find deceleration time using v²=u²+2as, calculate acceleration from given time/speeds, then use total distance constraint. The algebraic manipulation is straightforward, making this slightly easier than average for M1. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Shape 'V' | B1 | |
| Shape for last 22s (with \(V > 15\)) | B1 | |
| Figures: 15, 5, \(t\), 16, 22 on axes | B1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{1}{2}(15+5) \times t = 120\) | M1 | |
| \(\Rightarrow t = 12 \rightarrow T = 12 + 16 + 22 = 50\text{ s}\) | M1 A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(120 + \frac{1}{2}(V+5) \cdot 16 + 22V = 1000\) | M1 B1 A1 | |
| Solve: \(30V = 840 \Rightarrow V = 28\) | DM1 A1 (5) | Total: 11 |
## Question 3:
### Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| Shape 'V' | B1 | |
| Shape for last 22s (with $V > 15$) | B1 | |
| Figures: 15, 5, $t$, 16, 22 on axes | B1 (3) | |
### Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{1}{2}(15+5) \times t = 120$ | M1 | |
| $\Rightarrow t = 12 \rightarrow T = 12 + 16 + 22 = 50\text{ s}$ | M1 A1 (3) | |
### Part (c)
| Working | Marks | Notes |
|---------|-------|-------|
| $120 + \frac{1}{2}(V+5) \cdot 16 + 22V = 1000$ | M1 B1 A1 | |
| Solve: $30V = 840 \Rightarrow V = 28$ | DM1 A1 (5) | **Total: 11** |
---3. A car moves along a horizontal straight road, passing two points $A$ and $B$. At $A$ the speed of the car is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the driver passes $A$, he sees a warning sign $W$ ahead of him, 120 m away. He immediately applies the brakes and the car decelerates with uniform deceleration, reaching $W$ with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At $W$, the driver sees that the road is clear. He then immediately accelerates the car with uniform acceleration for 16 s to reach a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 } ( V > 15 )$. He then maintains the car at a constant speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Moving at this constant speed, the car passes $B$ after a further 22 s .
\begin{enumerate}[label=(\alph*)]
\item Sketch, in the space below, a speed-time graph to illustrate the motion of the car as it moves from $A$ to $B$.
\item Find the time taken for the car to move from $A$ to $B$.
The distance from $A$ to $B$ is 1 km .
\item Find the value of $V$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q3 [11]}}