| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration problem requiring two integrations with given initial conditions (starting from rest at O). The algebra is clean with simple polynomial terms, and part (ii) involves solving a quartic that factors nicely. Slightly above average due to the two-step integration process, but this is a standard M1 technique with no conceptual surprises. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| (integration) | M1 | For integrating to obtain \(v(t)\) |
| \(v(t) = 0.025t^3 - 0.75t^2 + 5t \quad (+0)\) | A1 | |
| (integration) | M1 | For integrating to obtain \(s(t)\) |
| \(s(t) = 0.00625t^4 - 0.25t^3 + 2.5t^2 \quad (+0)\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| (set \(s = 0\), \(t \neq 0\)) | M1 | For setting \(s = 0\) (\(t\) not zero) in their attempt at \(s\) obtained using integration only |
| \(t^4 - 40t^3 + 400t^2 = 0 \Rightarrow t^2(t-20)^2 = 0\) | M1 | For attempting to solve a quartic equation for \(s = 0\) where \(s\) was obtained using integration only |
| Time taken is \(20\) s | A1 | [3] \(t = 20\) only |
## Question 4:
### Part (i)
| Working/Answer | Mark | Guidance |
|---|---|---|
| (integration) | M1 | For integrating to obtain $v(t)$ |
| $v(t) = 0.025t^3 - 0.75t^2 + 5t \quad (+0)$ | A1 | |
| (integration) | M1 | For integrating to obtain $s(t)$ |
| $s(t) = 0.00625t^4 - 0.25t^3 + 2.5t^2 \quad (+0)$ | A1 | **[4]** |
### Part (ii)
| Working/Answer | Mark | Guidance |
|---|---|---|
| (set $s = 0$, $t \neq 0$) | M1 | For setting $s = 0$ ($t$ not zero) in their attempt at $s$ obtained using integration only |
| $t^4 - 40t^3 + 400t^2 = 0 \Rightarrow t^2(t-20)^2 = 0$ | M1 | For attempting to solve a quartic equation for $s = 0$ where $s$ was obtained using integration only |
| Time taken is $20$ s | A1 | **[3]** $t = 20$ only |
---4 A particle $P$ moves in a straight line. At time $t$ seconds after starting from rest at the point $O$ on the line, the acceleration of $P$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 0.075 t ^ { 2 } - 1.5 t + 5$.\\
(i) Find an expression for the displacement of $P$ from $O$ in terms of $t$.\\
(ii) Hence find the time taken for $P$ to return to the point $O$.
\hfill \mbox{\textit{CAIE M1 2015 Q4 [7]}}