| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Distance from velocity-time graph |
| Difficulty | Easy -1.3 This is a straightforward M1 question testing basic interpretation of speed-time graphs. Parts (a) and (b) require simple recall of what constant gradient and horizontal lines mean (constant acceleration and constant speed), while part (c) involves calculating area under the graph using basic geometry (trapezium). No problem-solving or novel insight required—purely routine application of standard graph interpretation. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| \(50 = 2 \times 22.5 + \frac{1}{2}a \cdot 4 \Rightarrow a = 2.5 \text{ m s}^{-2}\) | M1 A1 | Valid attempt at area of trapezoid. Area as trapezoid + rectangle (i.e. correct formula used with at most a slip in numbers). |
| \(v^2 = 22.5^2 + 2 \times 2.5 \times 100 \Rightarrow v \approx 31.7(2) \text{ m s}^{-1}\) | M1 A1∇ A1 | |
| \(v_B = 22.5 + 2 \times 2.5 = 27.5\) (must be used) | M1 | |
| \(31.72 = 27.5 + 2.5t\) OR \(50 = 27.5t + \frac{1}{2} \times 2.5t^2\) OR \(50 = \frac{1}{2}(27.5 + 31.72)t \Rightarrow t \approx 1.69 \text{ s}\) | M1 A1∇ A1 | (4) |
| OR \(31.72 = 22.5 + 2.5T\) OR \(100 = 22.5t + \frac{1}{2} \times 2.5T^2 \Rightarrow T \approx 3.69 \Rightarrow t \approx 3.69 - 2 = 1.69 \text{ s}\) | M1 A1∇ | (4) |
| OR \(50 = 31.7t - \frac{1}{2} \times 2.5t^2\) Solve quadratic to get \(t = 1.69 \text{ s}\) | M2 A1∇ | A1 (4) |
| $50 = 2 \times 22.5 + \frac{1}{2}a \cdot 4 \Rightarrow a = 2.5 \text{ m s}^{-2}$ | M1 A1 | Valid attempt at area of trapezoid. Area as trapezoid + rectangle (i.e. correct formula used with at most a slip in numbers). |
| $v^2 = 22.5^2 + 2 \times 2.5 \times 100 \Rightarrow v \approx 31.7(2) \text{ m s}^{-1}$ | M1 A1∇ A1 | |
| $v_B = 22.5 + 2 \times 2.5 = 27.5$ (must be used) | M1 | |
| $31.72 = 27.5 + 2.5t$ OR $50 = 27.5t + \frac{1}{2} \times 2.5t^2$ OR $50 = \frac{1}{2}(27.5 + 31.72)t \Rightarrow t \approx 1.69 \text{ s}$ | M1 A1∇ A1 | (4) |
| **OR** $31.72 = 22.5 + 2.5T$ OR $100 = 22.5t + \frac{1}{2} \times 2.5T^2 \Rightarrow T \approx 3.69 \Rightarrow t \approx 3.69 - 2 = 1.69 \text{ s}$ | M1 A1∇ | (4) |
| **OR** $50 = 31.7t - \frac{1}{2} \times 2.5t^2$ Solve quadratic to get $t = 1.69 \text{ s}$ | M2 A1∇ | A1 (4) |
**Guidance:** M1 for valid use of data (e.g. finding speed at $B$ by spurious means and using this to get $v$ at $C$ is M0). Accept answer as AWRT 31.7. M1 + M1 to get to an equation in the required $t$ (normally two stages, but they can do it in one via 3rd alternative above). Answer is cao. Hence premature approx (→ e.g. 1.68) is A0. But if they use a 3 s.f. answer from (b) and then give answer to (c) as 1.7, allow full marks. And accept 2 or 3 s.f. answer or better to (c).
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Figure 1\\
\includegraphics[max width=\textwidth, alt={}, center]{3a8395fd-6e44-48a1-8c97-3365a284956a-02_404_755_312_577}
Figure 1 shows the speed-time graph of a cyclist moving on a straight road over a 7 s period. The sections of the graph from $t = 0$ to $t = 3$, and from $t = 3$ to $t = 7$, are straight lines. The section from $t = 3$ to $t = 7$ is parallel to the $t$-axis.
State what can be deduced about the motion of the cyclist from the fact that
\begin{enumerate}[label=(\alph*)]
\item the graph from $t = 0$ to $t = 3$ is a straight line,
\item the graph from $t = 3$ to $t = 7$ is parallel to the $t$-axis.
\item Find the distance travelled by the cyclist during this 7 s period.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2006 Q1 [6]}}