| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a standard variable acceleration question requiring integration of a linear acceleration function with given initial conditions. The steps are routine: integrate to find velocity, apply initial conditions, solve for specific times, and integrate again for distance. While it involves multiple parts and careful handling of direction changes, it follows a well-practiced template with no novel problem-solving required. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \int(5.4 - 1.62t)\,dt\) | M1 | For using integration of \(a\) to find \(v\) |
| \(v = 5.4t - 0.81t^2\ (+C)\) | A1 | |
| \(5.4t - 0.81t^2 = 0\) | M1 | For solving \(v = 0\) |
| \(t = 6\frac{2}{3} = \frac{20}{3}\text{ s}\) | A1 |
| Answer | Marks |
|---|---|
| \(v(10) = -27\text{ ms}^{-1}\) | B1 |
| Inverted parabola | B1 |
| \(v = 0\) at \(t = 0\), negative at \(t = 10\) and through \(\left(6\frac{2}{3}, 0\right)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \int\left(5.4t - 0.81t^2\right)dt\) | M1 | For using integration of \(v\) to find \(s\) |
| \(s = 2.7t^2 - 0.27t^3\ (+C)\) | A1 | |
| At \(t = 6\frac{2}{3}\), displacement \(= 40\) | M1 | For evaluating the integral at the time when \(v = 0\) |
| At \(t = 10\), displacement \(= 0\) | M1 | For evaluating the integral at time \(t = 10\) |
| Total distance \(= 80\text{ m}\) | A1 |
## Question 7(i):
| $v = \int(5.4 - 1.62t)\,dt$ | M1 | For using integration of $a$ to find $v$ |
|---|---|---|
| $v = 5.4t - 0.81t^2\ (+C)$ | A1 | |
| $5.4t - 0.81t^2 = 0$ | M1 | For solving $v = 0$ |
| $t = 6\frac{2}{3} = \frac{20}{3}\text{ s}$ | A1 | |
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## Question 7(ii):
| $v(10) = -27\text{ ms}^{-1}$ | B1 | |
|---|---|---|
| Inverted parabola | B1 | |
| $v = 0$ at $t = 0$, negative at $t = 10$ and through $\left(6\frac{2}{3}, 0\right)$ | B1 | |
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## Question 7(iii):
| $s = \int\left(5.4t - 0.81t^2\right)dt$ | M1 | For using integration of $v$ to find $s$ |
|---|---|---|
| $s = 2.7t^2 - 0.27t^3\ (+C)$ | A1 | |
| At $t = 6\frac{2}{3}$, displacement $= 40$ | M1 | For evaluating the integral at the time when $v = 0$ |
| At $t = 10$, displacement $= 0$ | M1 | For evaluating the integral at time $t = 10$ |
| Total distance $= 80\text{ m}$ | A1 | |7 A particle moves in a straight line starting from rest from a point $O$. The acceleration of the particle at time $t \mathrm {~s}$ after leaving $O$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where
$$a = 5.4 - 1.62 t$$
(i) Find the positive value of $t$ at which the velocity of the particle is zero, giving your answer as an exact fraction.\\
(ii) Find the velocity of the particle at $t = 10$ and sketch the velocity-time graph for the first ten seconds of the motion.\\
(iii) Find the total distance travelled during the first ten seconds of the motion.\\
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 [12]}}