| 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 standard M1 variable acceleration question requiring integration to find distance, continuity conditions, and substitution. The piecewise function adds mild complexity, but all steps follow routine procedures: integrate v to get distance (5 marks suggests showing working), use continuity at t=15 to find B, then apply the distance formula for t≥15. Slightly above average due to the piecewise nature and multi-step reasoning, but no novel insight required. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For integrating \(v_1\) to find \(s_1\) | |
| \(\int_0^{15} v_1 dt = 225\) ➜ | A1 | |
| \(4[(15^2/2 − 0.05 \times 15^3/3) − (0 − 0)] = 225\) | 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] |
| Answer | Marks | Guidance |
|---|---|---|
| \(s_2(t) = Bt^{-1}(−1) + C\) | B1 | |
| \([−3375/15 + C = 225]\) | M1 | For using \(s_2(15) = 225\) to find C |
| Distance travelled is [450 − 3375/t] m (for t ⩾ 15) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \([450 − 3375/t = 315]\) | 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 ms⁻¹ | A1 | |
| [3] |
| Answer | Marks |
|---|---|
| \(s = \int_{15}^t 3375t^{-2}dt = −3375(\frac{1}{t} − \frac{1}{15})\) | B1 |
| \(= 225 − 3375/t\) | |
| Distance travelled = 225 + (225 − 3375/t) | M1 |
| Distance travelled is [450 − 3375/t] m (for t ⩾ 15) | A1 |
## (i)
| M1 | For integrating $v_1$ to find $s_1$
$\int_0^{15} v_1 dt = 225$ ➜ | A1 |
$4[(15^2/2 − 0.05 \times 15^3/3) − (0 − 0)] = 225$ | 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 ⩾ 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 ms⁻¹ | A1 |
| | [3]
# Question 7(ii) Alternative
$s = \int_{15}^t 3375t^{-2}dt = −3375(\frac{1}{t} − \frac{1}{15})$ | B1 |
$= 225 − 3375/t$ | |
Distance travelled = 225 + (225 − 3375/t) | M1 |
Distance travelled is [450 − 3375/t] m (for t ⩾ 15) | A1 |A vehicle is moving in a straight line. The velocity $v \text{ m s}^{-1}$ at time $t \text{ s}$ after the vehicle starts is given by
$$v = A(t - 0.05t^2) \text{ for } 0 \leq t \leq 15,$$
$$v = \frac{B}{t} \text{ for } t \geq 15,$$
where $A$ and $B$ are constants. The distance travelled by the vehicle between $t = 0$ and $t = 15$ is $225 \text{ m}$.
\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 $315 \text{ m}$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2010 Q7 [11]}}