| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.8 This question requires integration of piecewise functions, continuity conditions at t=60, and solving equations involving square roots. While the individual calculus steps are standard A-level, coordinating multiple constraints (initial conditions, distance traveled, continuity) across two different function forms and then working backwards from distance to speed requires careful multi-step reasoning beyond routine exercises. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = k_1t^2/2 - 0.005t^3/3 + C\) | M1 | For using \(s = \int v \, dt\) |
| \([k_1(60^2/2) - 0.005(60^3/3) = 540]\) | DM1 | For using limits 0 and 60 and equating to 540 |
| \(k_1 = 0.5\) | A1 | |
| \(0.5 \times 60 - 0.005 \times 60^2 = k_2 \div \sqrt{60}\) | M1 | For using \(v_1(60) = v_2(60)\) |
| \(k_2 = 12\sqrt{60}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = 540 + 12\sqrt{60}\int_{60}^{t} t^{-1/2} \, dt\) | M1 | For using \(s = 540 + 12\sqrt{60}\int_{60}^{t}(t^{-1/2})\,dt\) |
| \(s = 540 + 12\sqrt{60}(2\sqrt{t} - 2\sqrt{60}) = 24\sqrt{(60t)} - 900\) | A1 | Accept any other correct form for \(s\) if it is used in (iii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([24\sqrt{(60t)} - 900 = 1260]\) | M1 | For solving \(s(t) = 1260\) for \(t\) |
| \(t = 135\) | A1 | |
| \(v = 12\sqrt{60} \div \sqrt{135} \rightarrow \text{speed is } 8 \text{ ms}^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([16^2 = 24^2 + 2 \times 500a]\) | M1 | For using \(v^2 = u^2 + 2as\) |
| \(a = -0.32 \text{ ms}^{-2}\) | A1 | |
| M1 | For using Newton's second law | |
| \(5800 - 4800 - 12500g \times (h \div 500) = 12500(-0.32)\) | A1 | |
| Height of C is 20 m | A1 |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = k_1t^2/2 - 0.005t^3/3 + C$ | M1 | For using $s = \int v \, dt$ |
| $[k_1(60^2/2) - 0.005(60^3/3) = 540]$ | DM1 | For using limits 0 and 60 and equating to 540 |
| $k_1 = 0.5$ | A1 | |
| $0.5 \times 60 - 0.005 \times 60^2 = k_2 \div \sqrt{60}$ | M1 | For using $v_1(60) = v_2(60)$ |
| $k_2 = 12\sqrt{60}$ | A1 | AG |
**Total: 5 marks**
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 540 + 12\sqrt{60}\int_{60}^{t} t^{-1/2} \, dt$ | M1 | For using $s = 540 + 12\sqrt{60}\int_{60}^{t}(t^{-1/2})\,dt$ |
| $s = 540 + 12\sqrt{60}(2\sqrt{t} - 2\sqrt{60}) = 24\sqrt{(60t)} - 900$ | A1 | Accept any other correct form for $s$ if it is used in **(iii)** |
**Total: 2 marks**
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[24\sqrt{(60t)} - 900 = 1260]$ | M1 | For solving $s(t) = 1260$ for $t$ |
| $t = 135$ | A1 | |
| $v = 12\sqrt{60} \div \sqrt{135} \rightarrow \text{speed is } 8 \text{ ms}^{-1}$ | B1 | |
**Total: 3 marks**
---
# Previous Question (Part ii - Alternative):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[16^2 = 24^2 + 2 \times 500a]$ | M1 | For using $v^2 = u^2 + 2as$ |
| $a = -0.32 \text{ ms}^{-2}$ | A1 | |
| | M1 | For using Newton's second law |
| $5800 - 4800 - 12500g \times (h \div 500) = 12500(-0.32)$ | A1 | |
| Height of C is 20 m | A1 | |
**Total: 5 marks**7 A vehicle starts from rest at a point $O$ and moves in a straight line. Its speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds after leaving $O$ is defined as follows.
$$\begin{aligned}
\text { For } 0 & \leqslant t \leqslant 60 , \quad v = k _ { 1 } t - 0.005 t ^ { 2 } \\
\text { for } t \geqslant 60 , \quad v & = \frac { k _ { 2 } } { \sqrt { } t }
\end{aligned}$$
The distance travelled by the vehicle during the first 60 s is 540 m .\\
(i) Find the value of the constant $k _ { 1 }$ and show that $k _ { 2 } = 12 \sqrt { } ( 60 )$.\\
(ii) Find an expression in terms of $t$ for the total distance travelled when $t \geqslant 60$.\\
(iii) Find the speed of the vehicle when it has travelled a total distance of 1260 m .
\hfill \mbox{\textit{CAIE M1 2013 Q7 [10]}}