| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Standard +0.3 This is a straightforward kinematics question requiring differentiation of a given displacement function, solving a cubic equation that factorizes easily, and computing distance and average speed. All steps are routine calculus and mechanics procedures with no conceptual challenges beyond standard M1 content. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time taken \(= \frac{0.08}{0.0002} = 400\ \text{s}\) | B1 | |
| \(v = \frac{dx}{dt} = 0.16t - 0.0006t^2\) | B1 | |
| speed \(= -0.16\times400 + 0.0006\times400^2\) | M1 | For evaluating \(\pm v(400)\) |
| Speed at O is \(32\ \text{ms}^{-1}\) | A1 | Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Time to furthest point is \(0.16/0.0006\ \text{s}\) | B1↓ | \(v = 0.16t - kt^2\) or \(v = kt - 0.0006t^2\) from part (i) |
| \([0.08(800/3)^2 - 0.0002(800/3)^3]\ (\times2)\) | M1* | For evaluating \(x(t_{\text{furthest point}})\ (\times2)\) |
| Distance moved is 3790 m | A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| speed \(= 3790/400\ \text{ms}^{-1}\) | dM1* | For using 'average speed = total distance moved/time taken' |
| Average speed is \(9.48\ \text{ms}^{-1}\) | A1 | Total: 2 marks |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time taken $= \frac{0.08}{0.0002} = 400\ \text{s}$ | B1 | |
| $v = \frac{dx}{dt} = 0.16t - 0.0006t^2$ | B1 | |
| speed $= -0.16\times400 + 0.0006\times400^2$ | M1 | For evaluating $\pm v(400)$ |
| Speed at O is $32\ \text{ms}^{-1}$ | A1 | Total: 4 marks |
## Question 6(ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Time to furthest point is $0.16/0.0006\ \text{s}$ | B1↓ | $v = 0.16t - kt^2$ or $v = kt - 0.0006t^2$ from part (i) |
| $[0.08(800/3)^2 - 0.0002(800/3)^3]\ (\times2)$ | M1* | For evaluating $x(t_{\text{furthest point}})\ (\times2)$ |
| Distance moved is 3790 m | A1 | Total: 3 marks |
## Question 6(ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| speed $= 3790/400\ \text{ms}^{-1}$ | dM1* | For using 'average speed = total distance moved/time taken' |
| Average speed is $9.48\ \text{ms}^{-1}$ | A1 | Total: 2 marks |
---6 A particle $P$ starts from rest at a point $O$ of a straight line and moves along the line. The displacement of the particle at time $t \mathrm {~s}$ after leaving $O$ is $x \mathrm {~m}$, where
$$x = 0.08 t ^ { 2 } - 0.0002 t ^ { 3 }$$
(i) Find the value of $t$ when $P$ returns to $O$ and find the speed of $P$ as it passes through $O$ on its return.\\
(ii) For the motion of $P$ until the instant it returns to $O$, find
\begin{enumerate}[label=(\alph*)]
\item the total distance travelled,
\item the average speed.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2015 Q6 [9]}}