| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.8 This is a straightforward velocity-time graph question requiring only basic interpretation: finding distances from areas under the graph (trapezium/triangle areas), reading time values directly, and applying a=Δv/Δt. All techniques are standard M1 material with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature and need for careful graph reading. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| \(+/-3x20/2\), 30 m | M1, A1 | Use area of scalene triangle(s). Not suvat. Accept -30 |
| Answer | Marks | Guidance |
|---|---|---|
| \((t+4)x3/2 = 30\) or \(3t/2 = 30 - 4x3\), \(t = 16\) or \(t = 12\), \(T = 76\) | M1, A1, A1, A1 | Equates scalene trapezium area to distance (i). [(T-60)+4]x3/2 = 30, award A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T(accdn) = 3/0.4 = 7.5\) s), \(decn = 3/(76-60] - 4 - 7.5)\), \(decn = (+/-) 2/3\) m s\(^{-2}\) OR \(S(accen) = 3^2/(2x0.4) = (11.25\) m), \(decn = 3^2/[2x(30 - 3x4 - 11.25)]\), \(decn = (+/-) 2/3\) m s\(^{-2}\) | B1, M1, A1, B1, M1, A1 | Or 3 = decn x (76-60] - 4 - 7.5). (+/-) 0.667 or better - accept 0.6 recurring |
**Part i**
| $+/-3x20/2$, 30 m | M1, A1 | Use area of scalene triangle(s). Not suvat. Accept -30 |
**Part ii**
| $(t+4)x3/2 = 30$ or $3t/2 = 30 - 4x3$, $t = 16$ or $t = 12$, $T = 76$ | M1, A1, A1, A1 | Equates scalene trapezium area to distance (i). [(T-60)+4]x3/2 = 30, award A2 |
**Part iii**
| $T(accdn) = 3/0.4 = 7.5$ s), $decn = 3/(76-60] - 4 - 7.5)$, $decn = (+/-) 2/3$ m s$^{-2}$ OR $S(accen) = 3^2/(2x0.4) = (11.25$ m), $decn = 3^2/[2x(30 - 3x4 - 11.25)]$, $decn = (+/-) 2/3$ m s$^{-2}$ | B1, M1, A1, B1, M1, A1 | Or 3 = decn x (76-60] - 4 - 7.5). (+/-) 0.667 or better - accept 0.6 recurring |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{4b703cf9-b3d3-4210-b57b-89136595f8a5-03_508_1397_255_374}
The diagram shows the ( $t , v$ ) graph for a lorry delivering waste to a recycling centre. The graph consists of six straight line segments. The lorry reverses in a straight line from a stationary position on a weighbridge before coming to rest. It deposits its waste and then moves forwards in a straight line accelerating to a maximum speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It maintains this speed for 4 s and then decelerates, coming to rest at the weighbridge.\\
(i) Calculate the distance from the weighbridge to the point where the lorry deposits the waste.\\
(ii) Calculate the time which elapses between the lorry leaving the weighbridge and returning to it.\\
(iii) Given that the acceleration of the lorry when it is moving forwards is $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, calculate its final deceleration.
\hfill \mbox{\textit{OCR M1 2010 Q5 [9]}}