| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration question requiring integration to find distance, continuity conditions at t=15, and working backwards from distance to speed. All techniques are standard M1 material with clear signposting across three routine parts, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For integrating \(v_1\) to find \(s_1\) |
| \(\int_0^{15} v_1 \text{d}t = 225 \Rightarrow\) | A1 | |
| \(A[(15^2/2 - 0.05 \times 15^3/3) - (0 - 0)] = 225\) | A1 | |
| \(A = 4\) | A1 | |
| \([4(15 - 0.05 \times 15^2) = B/15^2]\) | M1 | For using \(v_1(15) = v_2(15)\) |
| \(B = 3375\) | A1 | AG |
| [5] | ||
| (ii) \(s_2(t) = Bt^{-1}(-1) + C\) | B1 | |
| \([-3375/15 + C = 225]\) | M1 | For using \(s_2(15) = 225\) to find C |
| Distance travelled is \([450 - 3375/t]\) m (for \(t \geq 15\)) | A1 | |
| [3] | ||
| (iii) \([450 - 3375/t = 315]\) | M1 | For attempting to solve \(s_1(t) = 315\) |
| \([v = 3375/25^2]\) | M1 | For substituting in \(v = 3375/t^2\) |
| Speed is \(5.4 \text{ ms}^{-1}\) | A1 | |
| [3] |
| Answer | Marks |
|---|---|
| \(s = \int_{15}^t 3375t^{-2} \text{d}t = -3375(\frac{1}{t} - \frac{1}{15})\) | B1 |
| \(= 225 - 3375/t\) | |
| Distance travelled \(= 225 + (225 - 3375/t)\) | M1 |
| Distance travelled is \([450 - 3375/t]\) m (for \(t \geq 15\)) | A1 |
**(i)** | M1 | For integrating $v_1$ to find $s_1$
$\int_0^{15} v_1 \text{d}t = 225 \Rightarrow$ | A1 |
$A[(15^2/2 - 0.05 \times 15^3/3) - (0 - 0)] = 225$ | A1 |
$A = 4$ | A1 |
$[4(15 - 0.05 \times 15^2) = B/15^2]$ | M1 | For using $v_1(15) = v_2(15)$
$B = 3375$ | A1 | AG
| | | **[5]**
**(ii)** $s_2(t) = Bt^{-1}(-1) + C$ | B1 |
$[-3375/15 + C = 225]$ | M1 | For using $s_2(15) = 225$ to find C
Distance travelled is $[450 - 3375/t]$ m (for $t \geq 15$) | A1 |
| | | **[3]**
**(iii)** $[450 - 3375/t = 315]$ | M1 | For attempting to solve $s_1(t) = 315$
$[v = 3375/25^2]$ | M1 | For substituting in $v = 3375/t^2$
Speed is $5.4 \text{ ms}^{-1}$ | A1 |
| | | **[3]**
---
## Question 7 Alternative for (ii)
$s = \int_{15}^t 3375t^{-2} \text{d}t = -3375(\frac{1}{t} - \frac{1}{15})$ | B1 |
$= 225 - 3375/t$ | |
Distance travelled $= 225 + (225 - 3375/t)$ | M1 |
Distance travelled is $[450 - 3375/t]$ m (for $t \geq 15$) | A1 |A vehicle is moving in a straight line. The velocity $v$ m s$^{-1}$ at time $t$ s after the vehicle starts is given by
$$v = A(t - 0.05t^2) \quad \text{for } 0 \leq t \leq 15,$$
$$v = \frac{B}{t^2} \quad \text{for } t \geq 15,$$
where $A$ and $B$ are constants. The distance travelled by the vehicle between $t = 0$ and $t = 15$ is 225 m.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $A$ and show that $B = 3375$. [5]
\item Find an expression in terms of $t$ for the total distance travelled by the vehicle when $t \geq 15$. [3]
\item Find the speed of the vehicle when it has travelled a total distance of 315 m. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q7 [11]}}