| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a standard variable acceleration question requiring integration with initial conditions. Students integrate acceleration to find velocity, set derivative to zero for maximum, and integrate again for distance. The piecewise nature and fractional power in part (iii) add slight complexity, but the techniques are routine for M1 level with clear signposting of required methods. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration \(= 0\) when \(t = 5\) from \(25 - t^2 = 0\) | B1 | |
| \([v = 25t - \frac{1}{3}t^3]\) | M1 | Use of integration |
| \([\text{Max speed} = 25 \times 5 - \frac{1}{3} \times 5^3]\) | M1 | Substitution for \(t\) |
| Max speed \(= 83\frac{1}{3} \text{ ms}^{-1}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([s = 12\frac{1}{2}t^2 - \frac{1}{12}t^4]\) | M1 | Use of integration |
| Distance \(= 260\) m (260.4166...) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(t=9\), \(v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18\) | B1ft | ft \(v\) from (i) |
| \(\left[s = \int_9^{25}\left(-3t^{-\frac{1}{2}}\right)dt = \left[-6t^{\frac{1}{2}}\right]\right]\) | M1 | Use of integration |
| \(\left[\text{Change in velocity from } t=9 \text{ to } t=25 = \left[-6t^{\frac{1}{2}}\right] = -6\times5 + 6\times3 = -12\right]\) | M1 | Substituting limits |
| Velocity at \(t = 25\) is \(-18 - 12 = -30 \text{ ms}^{-1}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(t=9\), \(v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18\) | B1ft | ft \(v\) from (i) |
| \([s = \int -3t^{-\frac{1}{2}}\,dt = -6t^{\frac{1}{2}} (+C)]\) | M1 | Use of integration |
| \([t=9, v=-18 \rightarrow C=0,\ t=25,\ v = -6 \times 25^{\frac{1}{2}}]\) | M1 | Finds \(C\) and substitutes \(t=25\) |
| Velocity at \(t = 25\) is \(-30 \text{ ms}^{-1}\) | A1 | |
| 4 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration $= 0$ when $t = 5$ from $25 - t^2 = 0$ | B1 | |
| $[v = 25t - \frac{1}{3}t^3]$ | M1 | Use of integration |
| $[\text{Max speed} = 25 \times 5 - \frac{1}{3} \times 5^3]$ | M1 | Substitution for $t$ |
| Max speed $= 83\frac{1}{3} \text{ ms}^{-1}$ | A1 | |
| | 4 | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[s = 12\frac{1}{2}t^2 - \frac{1}{12}t^4]$ | M1 | Use of integration |
| Distance $= 260$ m (260.4166...) | A1 | |
| | 2 | |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $t=9$, $v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18$ | B1ft | ft $v$ from (i) |
| $\left[s = \int_9^{25}\left(-3t^{-\frac{1}{2}}\right)dt = \left[-6t^{\frac{1}{2}}\right]\right]$ | M1 | Use of integration |
| $\left[\text{Change in velocity from } t=9 \text{ to } t=25 = \left[-6t^{\frac{1}{2}}\right] = -6\times5 + 6\times3 = -12\right]$ | M1 | Substituting limits |
| Velocity at $t = 25$ is $-18 - 12 = -30 \text{ ms}^{-1}$ | A1 | |
| | 4 | |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $t=9$, $v = 25 \times 9 - \frac{1}{3} \times 9^3 = -18$ | B1ft | ft $v$ from (i) |
| $[s = \int -3t^{-\frac{1}{2}}\,dt = -6t^{\frac{1}{2}} (+C)]$ | M1 | Use of integration |
| $[t=9, v=-18 \rightarrow C=0,\ t=25,\ v = -6 \times 25^{\frac{1}{2}}]$ | M1 | Finds $C$ and substitutes $t=25$ |
| Velocity at $t = 25$ is $-30 \text{ ms}^{-1}$ | A1 | |
| | 4 | |7 A particle moves in a straight line. The particle is initially at rest at a point $O$ on the line. At time $t \mathrm {~s}$ after leaving $O$, the acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of the particle is given by $a = 25 - t ^ { 2 }$ for $0 \leqslant t \leqslant 9$.\\
(i) Find the maximum velocity of the particle in this time period.\\
(ii) Find the total distance travelled until the maximum velocity is reached.\\
The acceleration of the particle for $t > 9$ is given by $a = - 3 t ^ { - \frac { 1 } { 2 } }$.\\
(iii) Find the velocity of the particle when $t = 25$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M1 2018 Q7 [10]}}