| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| 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 SUVAT and travel graph question where all necessary information is provided explicitly. Students must apply basic kinematic formulas (a = v/t), calculate areas under velocity-time graphs (trapeziums/rectangles), and perform simple arithmetic to find the time in the shop. It requires no problem-solving insight—just methodical application of standard M1 techniques with clearly defined stages. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial acceleration \(= \frac{15}{6} = 2.5\) m s\(^{-2}\) | B1 | |
| Final deceleration: from graph decelerates from 15 to 0 in 2 s \(= \frac{15}{2} = 7.5\) m s\(^{-2}\) | B1 | |
| (deceleration of car shown as \(-2\) to some value on graph) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance = area under car's motion portion | M1 | |
| \(= \frac{1}{2}(6)(15) + 11(15) + \frac{1}{2}(2)(15)\) | M1 | |
| \(= 45 + 165 + 15 = 225\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driver walks back: \(20 \times 2 = 40\) m in 20 s | M1 | |
| Driver jogs to car at 5 m s\(^{-1}\); car is 225 m from shop | M1 | |
| Time jogging: \(\frac{225 - 40}{5}\) ... set up equation using total time = 60 s | M1 | |
| Time in shop \(= 60 - 6 - 11 - 2 - 20 - \frac{185}{5}\) | M1 | |
| \(= 60 - 39 - 37 = ...\); shop time \(= 4\) s | A1 |
# Question 4:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial acceleration $= \frac{15}{6} = 2.5$ m s$^{-2}$ | B1 | |
| Final deceleration: from graph decelerates from 15 to 0 in 2 s $= \frac{15}{2} = 7.5$ m s$^{-2}$ | B1 | |
| (deceleration of car shown as $-2$ to some value on graph) | B1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance = area under car's motion portion | M1 | |
| $= \frac{1}{2}(6)(15) + 11(15) + \frac{1}{2}(2)(15)$ | M1 | |
| $= 45 + 165 + 15 = 225$ m | A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driver walks back: $20 \times 2 = 40$ m in 20 s | M1 | |
| Driver jogs to car at 5 m s$^{-1}$; car is 225 m from shop | M1 | |
| Time jogging: $\frac{225 - 40}{5}$ ... set up equation using total time = 60 s | M1 | |
| Time in shop $= 60 - 6 - 11 - 2 - 20 - \frac{185}{5}$ | M1 | |
| $= 60 - 39 - 37 = ...$; shop time $= 4$ s | A1 | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{ce4c43e6-da4f-4c02-ab0f-01a21717949c-2_657_1495_1539_324}
A car travelling on a straight road accelerates from rest to a speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 6 s . It continues at constant speed for 11 s and then decelerates to rest in 2 s . The driver gets out of the car and walks at a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 20 s back to a shop which he enters. Some time later he leaves the shop and jogs to the car at a speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. He arrives at the vehicle 60 s after it began to accelerate from rest. The diagram, which has six straight line segments, shows the $( t , v )$ graph for the motion of the driver.\\
(i) Calculate the initial acceleration and final deceleration of the car.\\
(ii) Calculate the distance the car travels.\\
(iii) Calculate the length of time the driver is in the shop.
\hfill \mbox{\textit{OCR M1 2011 Q4 [10]}}