| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring differentiation to find acceleration, substitution of given conditions to form simultaneous equations, and integration to find distance. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = \dfrac{9}{4} + \dfrac{b}{(5+1)^2} - c \times 5^2\) | B1 | Use of \(v(5) = 0\) to form equation in \(b\) and \(c\) |
| \(a = -2\dfrac{b}{(t+1)^3} - 2ct\) | M1 | For use of \(a = \dfrac{dv}{dt}\) and \(a(5) = -\dfrac{13}{12}\) |
| \(-\dfrac{13}{12} = -2\dfrac{b}{(5+1)^3} - 2c \times 5\) | A1 | |
| \(\dfrac{b}{36} - 25c = -\dfrac{9}{4}\) and \(-\dfrac{b}{108} - 10c = -\dfrac{13}{12}\) leading to \(b = \ldots\) or \(c = \ldots\) | M1 | Attempts to solve simultaneous equations |
| \(b = 9\) and \(c = 0.1\) | A1 | \(b = 9\) (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\displaystyle\int\left(\dfrac{9}{4} + \dfrac{9}{(t+1)^2} - 0.1t^2\right)dt = \ldots\) | M1 | For use of \(s = \int v \, dt\) |
| \(= \dfrac{9}{4}t - \dfrac{9}{(t+1)} - \dfrac{1}{30}t^3 [+K]\) | A1 FT | FT their value of \(c\) from (a) |
| \(\left[\dfrac{9}{4}t - \dfrac{9}{(t+1)} - \dfrac{1}{30}t^3\right]_5^{10} = \left[\left(\dfrac{9}{4}\times 10 - \dfrac{9}{(10+1)} - \dfrac{1}{30}\times 10^3\right) - \left(\dfrac{9}{4}\times 5 - \dfrac{9}{(5+1)} - \dfrac{1}{30}\times 5^3\right)\right]\) | M1 | For evaluation from 0 to 5 or from 5 to 10 |
| \(= 5.583 + 9 + 11.651 + 5.583\) or \(\dfrac{175}{12} + \dfrac{2275}{132}\) | M1 | For evaluation from 0 to 5 and from 5 to 10 to find distance travelled |
| \(= 31.8 \text{ m}\) | A1 | or \(\dfrac{350}{11}\) |
## Question 7:
### Part (a):
$0 = \dfrac{9}{4} + \dfrac{b}{(5+1)^2} - c \times 5^2$ | B1 | Use of $v(5) = 0$ to form equation in $b$ and $c$
$a = -2\dfrac{b}{(t+1)^3} - 2ct$ | M1 | For use of $a = \dfrac{dv}{dt}$ and $a(5) = -\dfrac{13}{12}$
$-\dfrac{13}{12} = -2\dfrac{b}{(5+1)^3} - 2c \times 5$ | A1 |
$\dfrac{b}{36} - 25c = -\dfrac{9}{4}$ and $-\dfrac{b}{108} - 10c = -\dfrac{13}{12}$ leading to $b = \ldots$ or $c = \ldots$ | M1 | Attempts to solve simultaneous equations
$b = 9$ and $c = 0.1$ | A1 | $b = 9$ (AG)
### Part (b):
$\displaystyle\int\left(\dfrac{9}{4} + \dfrac{9}{(t+1)^2} - 0.1t^2\right)dt = \ldots$ | M1 | For use of $s = \int v \, dt$
$= \dfrac{9}{4}t - \dfrac{9}{(t+1)} - \dfrac{1}{30}t^3 [+K]$ | A1 FT | FT their value of $c$ from (a)
$\left[\dfrac{9}{4}t - \dfrac{9}{(t+1)} - \dfrac{1}{30}t^3\right]_5^{10} = \left[\left(\dfrac{9}{4}\times 10 - \dfrac{9}{(10+1)} - \dfrac{1}{30}\times 10^3\right) - \left(\dfrac{9}{4}\times 5 - \dfrac{9}{(5+1)} - \dfrac{1}{30}\times 5^3\right)\right]$ | M1 | For evaluation from 0 to 5 or from 5 to 10
$= 5.583 + 9 + 11.651 + 5.583$ or $\dfrac{175}{12} + \dfrac{2275}{132}$ | M1 | For evaluation from 0 to 5 and from 5 to 10 to find distance travelled
$= 31.8 \text{ m}$ | A1 | or $\dfrac{350}{11}$7 A particle $P$ moves in a straight line through a point $O$. The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of $P$, at time $t \mathrm {~s}$ after passing $O$, is given by
$$v = \frac { 9 } { 4 } + \frac { b } { ( t + 1 ) ^ { 2 } } - c t ^ { 2 }$$
where $b$ and $c$ are positive constants. At $t = 5$, the velocity of $P$ is zero and its acceleration is $- \frac { 13 } { 12 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $b = 9$ and find the value of $c$.
\item Given that the velocity of $P$ is zero only at $t = 5$, find the distance travelled in the first 10 seconds of motion.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q7 [10]}}