| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise velocity function question requiring basic differentiation for acceleration, solving v=0 algebraically, and integration for distance. All techniques are standard M1 procedures with no conceptual challenges—slightly easier than average due to the routine nature of each part. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration \(= 0.4\) m s\(^{-2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{100}{t^2} - 0.1t = 0\) | M1 | For setting \(v = 0\) and solving for \(t\) |
| \(t = 10\) s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance \(t = 0\) to \(t = 5\) is \(\frac{1}{2}(1.5 + 3.5) \times 5 = 12.5\) | B1 | Trapezium rule or integration |
| \(s(t) = \int\!\left(\dfrac{100}{t^2} - 0.1t\right)dt\) | M1 | For integration |
| \(= -\dfrac{100}{t} - 0.05t^2 \;(+C)\) | A1 | Correct integration |
| \(s(10) - s(5)\) | M1 | Use limits 5 and 10 used or find \(+\,C\) |
| Total distance \(= 12.5 + 6.25 = 18.75\) m | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration $= 0.4$ m s$^{-2}$ | B1 | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{100}{t^2} - 0.1t = 0$ | M1 | For setting $v = 0$ and solving for $t$ |
| $t = 10$ s | A1 | |
---
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $t = 0$ to $t = 5$ is $\frac{1}{2}(1.5 + 3.5) \times 5 = 12.5$ | B1 | Trapezium rule or integration |
| $s(t) = \int\!\left(\dfrac{100}{t^2} - 0.1t\right)dt$ | M1 | For integration |
| $= -\dfrac{100}{t} - 0.05t^2 \;(+C)$ | A1 | Correct integration |
| $s(10) - s(5)$ | M1 | Use limits 5 and 10 used or find $+\,C$ |
| Total distance $= 12.5 + 6.25 = 18.75$ m | A1 | |
---5 A particle starts from a point $O$ and moves in a straight line. The velocity of the particle at time $t \mathrm {~s}$ after leaving $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where
$$\begin{array} { l l }
v = 1.5 + 0.4 t & \text { for } 0 \leqslant t \leqslant 5 , \\
v = \frac { 100 } { t ^ { 2 } } - 0.1 t & \text { for } t \geqslant 5 .
\end{array}$$
(i) Find the acceleration of the particle during the first 5 seconds of motion.\\
(ii) Find the value of $t$ when the particle is instantaneously at rest.\\
(iii) Find the total distance travelled by the particle in the first 10 seconds of motion.\\
\hfill \mbox{\textit{CAIE M1 2017 Q5 [8]}}