| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.8 This is a straightforward M1 question requiring basic SUVAT calculations (distance from trapezium area under graph), recognition that constant gradient implies constant acceleration/force, and application of F=ma with given deceleration. All steps are routine with no problem-solving insight needed, making it easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Distance} = \frac{1}{2}(30+17) \times 3 + 4 \times 17\) | M1 A1 M1 | |
| \(= 138.5 \text{ m}\) | A1 | (4 marks) |
| OR: \(\frac{1}{2} \times 3(30-17) + 3 \times 17 + 4 \times 17 = 138.5 \text{ m}\) | M1 A1 M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Straight line on \(v\)–\(t\) graph \(\Rightarrow\) constant deceleration | M1 | |
| \(F = ma \Rightarrow F\) constant | A1 cso | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Deceleration} = \frac{30-17}{3}\) | M1 | |
| \(\text{Force} = 1200 \times \left(\frac{30-17}{3}\right) \approx 5200 \text{ N}\) | M1 A1 | (3 marks) Total: 9 |
## Question 3:
### Part (a):
$\text{Distance} = \frac{1}{2}(30+17) \times 3 + 4 \times 17$ | M1 A1 M1 |
$= 138.5 \text{ m}$ | A1 | (4 marks)
**OR:** $\frac{1}{2} \times 3(30-17) + 3 \times 17 + 4 \times 17 = 138.5 \text{ m}$ | M1 A1 M1 A1 |
### Part (b):
Straight line on $v$–$t$ graph $\Rightarrow$ constant deceleration | M1 |
$F = ma \Rightarrow F$ constant | A1 cso | (2 marks)
### Part (c):
$\text{Deceleration} = \frac{30-17}{3}$ | M1 |
$\text{Force} = 1200 \times \left(\frac{30-17}{3}\right) \approx 5200 \text{ N}$ | M1 A1 | (3 marks) **Total: 9**
---3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{218383c1-0875-46f2-9416-8e827065a7a6-3_540_1223_348_455}
\end{center}
\end{figure}
A car of mass 1200 kg moves along a straight horizontal road. In order to obey a speed restriction, the brakes of the car are applied for 3 s , reducing the car's speed from $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The brakes are then released and the car continues at a constant speed of $17 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for a further 4 s . Figure 2 shows a sketch of the speed-time graph of the car during the 7 s interval. The graph consists of two straight line segments.
\begin{enumerate}[label=(\alph*)]
\item Find the total distance moved by the car during this 7 s interval.
\item Explain briefly how the speed-time graph shows that, when the brakes are applied, the car experiences a constant retarding force.
\item Find the magnitude of this retarding force.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q3 [9]}}