| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward non-constant acceleration problem requiring standard calculus techniques: factorizing a cubic to find when v=0, integrating velocity to find distance, and differentiating to find maximum velocity. All steps are routine applications of A-level methods with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t^2(0.009 - 0.0001t) = 0\) | M1 | For attempting to solve \(v(t) = 0\) for \(t \neq 0\) |
| Time is \(90 \text{ s}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For attempting to integrate \(v(t)\) | |
| \(s = 0.003t^3 - 0.000025t^4 \quad (+C)\) | A1 | |
| \((2187 - 1640.25) - (0 - 0)\) | M1 | For attempting to find \(s(\text{ans i}) - s(0)\) [the subtraction of \(s(0)\) is implied if \(C\) is found to be zero or if \(C\) is absent] |
| Distance is \(547 \text{ m}\) | A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.018t - 0.0003t^2 = 0 \rightarrow t(0.018 - 0.0003t) = 0\) | M1 | For obtaining \(\dot{v}\) and attempting to solve \(\dot{v} = 0\) |
| \(t = 60\) (may be implied) | A1 | |
| \(32.4 - 21.6\) | M1 | For attempting to find \(v(60)\) |
| Maximum speed is \(10.8 \text{ ms}^{-1}\) | A1 | 4 marks total |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t^2(0.009 - 0.0001t) = 0$ | M1 | For attempting to solve $v(t) = 0$ for $t \neq 0$ |
| Time is $90 \text{ s}$ | A1 | **2 marks total** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For attempting to integrate $v(t)$ |
| $s = 0.003t^3 - 0.000025t^4 \quad (+C)$ | A1 | |
| $(2187 - 1640.25) - (0 - 0)$ | M1 | For attempting to find $s(\text{ans i}) - s(0)$ [the subtraction of $s(0)$ is implied if $C$ is found to be zero or if $C$ is absent] |
| Distance is $547 \text{ m}$ | A1 | **4 marks total** |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.018t - 0.0003t^2 = 0 \rightarrow t(0.018 - 0.0003t) = 0$ | M1 | For obtaining $\dot{v}$ and attempting to solve $\dot{v} = 0$ |
| $t = 60$ (may be implied) | A1 | |
| $32.4 - 21.6$ | M1 | For attempting to find $v(60)$ |
| Maximum speed is $10.8 \text{ ms}^{-1}$ | A1 | **4 marks total** |7 A particle starts from rest at the point $A$ and travels in a straight line until it reaches the point $B$. The velocity of the particle $t$ seconds after leaving $A$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 0.009 t ^ { 2 } - 0.0001 t ^ { 3 }$. Given that the velocity of the particle when it reaches $B$ is zero, find\\
(i) the time taken for the particle to travel from $A$ to $B$,\\
(ii) the distance $A B$,\\
(iii) the maximum velocity of the particle.
\hfill \mbox{\textit{CAIE M1 2004 Q7 [10]}}