| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Standard +0.3 This is a straightforward mechanics question combining basic kinematics with simple calculus. Parts (i)-(ii) require only reading a velocity-time graph (area under graph, gradient). Parts (iii)-(iv) involve differentiating a quadratic velocity function and integrating to find displacement—standard M1 techniques with no problem-solving insight required. Slightly above average due to the multi-part nature and calculus application, but all steps are routine. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| For using the idea that area represents the distance travelled. | M1 | |
| For any two of \(\frac{1}{2} \times 100 \times 4.8\), \(\frac{1}{2} \times 200(4.8 + 7.2)\), \(\frac{1}{2} \times 200 \times 7.2\) (240, 1200, 720) | A1 | |
| Distance is 2160 m | A1 | 3 marks |
| For using the idea that the initial acceleration is the gradient of the first line segment or for using \(v = at\) \((4.8 = 100a)\) or \(v^2 = 2as\) \((4.8^2 = 2a \times 240)\) | M1 | |
| Acceleration is 0.048 m s\(^{-2}\) | A1 | 2 marks |
| \(a = 0.06 - 0.00024t\) | B1 | |
| Acceleration is greater by 0.012 m s\(^{-2}\) [√ for \(0.06 -\) ans(ii) (must be +ve) and/or wrong coefficient of t in a(t)] [Accept 'acceleration is 1.25 times greater'] | B1 √ | 2 marks |
| B's velocity is a maximum when \(0.06 - 0.00024t = 0\) | B1 √ | |
| [√ wrong coefficient of t in a(t)] | ||
| For the method of finding the area representing \(s_A(250)\) | M1 | |
| \(240 + \frac{1}{2}(4.8 + 6.6)150\) or \(240 + (4.8 \times 150 + \frac{1}{2} \times 0.012 \times 150^2)\) (1095) | A1 | |
| For using the idea that \(s_B\) is obtained from integration | M1 | |
| \(0.03t^2 - 0.00004t^3\) | A1 | |
| Required distance is 155 m (√ dependent on both M marks) | A1 √ | 6 marks |
For using the idea that area represents the distance travelled. | M1 |
For any two of $\frac{1}{2} \times 100 \times 4.8$, $\frac{1}{2} \times 200(4.8 + 7.2)$, $\frac{1}{2} \times 200 \times 7.2$ (240, 1200, 720) | A1 |
Distance is 2160 m | A1 | 3 marks
For using the idea that the initial acceleration is the gradient of the first line segment or for using $v = at$ $(4.8 = 100a)$ or $v^2 = 2as$ $(4.8^2 = 2a \times 240)$ | M1 |
Acceleration is 0.048 m s$^{-2}$ | A1 | 2 marks
$a = 0.06 - 0.00024t$ | B1 |
Acceleration is greater by 0.012 m s$^{-2}$ [√ for $0.06 -$ ans(ii) (must be +ve) and/or wrong coefficient of t in a(t)] [Accept 'acceleration is 1.25 times greater'] | B1 √ | 2 marks
B's velocity is a maximum when $0.06 - 0.00024t = 0$ | B1 √ |
[√ wrong coefficient of t in a(t)] | |
For the method of finding the area representing $s_A(250)$ | M1 |
$240 + \frac{1}{2}(4.8 + 6.6)150$ or $240 + (4.8 \times 150 + \frac{1}{2} \times 0.012 \times 150^2)$ (1095) | A1 |
For using the idea that $s_B$ is obtained from integration | M1 |
$0.03t^2 - 0.00004t^3$ | A1 |
Required distance is 155 m (√ dependent on both M marks) | A1 √ | 6 marks7\\
\includegraphics[max width=\textwidth, alt={}, center]{5cba3e17-3979-4c22-a415-2cdd60f09289-4_547_1237_269_456}
A tractor $A$ starts from rest and travels along a straight road for 500 seconds. The velocity-time graph for the journey is shown above. This graph consists of three straight line segments. Find\\
(i) the distance travelled by $A$,\\
(ii) the initial acceleration of $A$.
Another tractor $B$ starts from rest at the same instant as $A$, and travels along the same road for 500 seconds. Its velocity $t$ seconds after starting is $\left( 0.06 t - 0.00012 t ^ { 2 } \right) \mathrm { m } \mathrm { s } ^ { - 1 }$. Find\\
(iii) how much greater $B$ 's initial acceleration is than $A$ 's,\\
(iv) how much further $B$ has travelled than $A$, at the instant when $B$ 's velocity reaches its maximum.
\hfill \mbox{\textit{CAIE M1 2003 Q7 [13]}}