| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.8 This question requires integration of fractional powers, finding when velocity equals zero by solving a quadratic, determining maximum speed by setting acceleration to zero, and computing displacement with careful attention to initial conditions. While the techniques are standard M1 content, the fractional indices and multi-step nature with several conceptual transitions (acceleration→velocity→displacement, identifying maximum speed occurs when a=0) make this moderately challenging, above average difficulty. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For an attempt at integration | \*M1 | Power of at least one term increased by 1 and the coefficient changed |
| \(v = 4t^{\frac{1}{2}} - \frac{2}{5}t^{\frac{3}{2}} [+C]\) | A1 | Correct \(v\) |
| \(4t^{\frac{1}{2}} - \frac{2}{5}t^{\frac{3}{2}} = 0\) | DM1 | Equating *their* 2-term \(v\) to zero and attempt to solve for \(t\) or \(k\) |
| \(k = 10\) | A1 | Final answer \(t=10\) is A0 |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Max speed when \(2t^{-\frac{1}{2}} - \frac{3}{5}t^{\frac{1}{2}} = 0\) | M1 | Attempt to solve \(a=0\) and find a value of \(t\) |
| \(t = \frac{10}{3}\) | A1 | |
| Maximum speed \(= 4.87\text{ ms}^{-1}\) to 3 sf | B1 | Allow maximum speed as \(\frac{8\sqrt{30}}{9}\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt at integration of their \(v\) | \*M1 | Power of at least one term increased by 1 and coefficient changed |
| \(s = \frac{8}{3}t^{\frac{3}{2}} - \frac{4}{25}t^{\frac{5}{2}} [+C]\) | A1 | Correct \(s\) |
| Substitute their \(t = \frac{10}{3}\) and \(t = 10\) | DM1 | Use their \(t = 10\) and their \(t = \frac{10}{3} (\neq 0)\) correctly |
| Distance \(= 20.7\) m | A1 | Distance \(= 20.7479...\) |
## Question 6:
### Part 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| For an attempt at integration | \*M1 | Power of at least one term increased by 1 and the coefficient changed |
| $v = 4t^{\frac{1}{2}} - \frac{2}{5}t^{\frac{3}{2}} [+C]$ | A1 | Correct $v$ |
| $4t^{\frac{1}{2}} - \frac{2}{5}t^{\frac{3}{2}} = 0$ | DM1 | Equating *their* 2-term $v$ to zero and attempt to solve for $t$ or $k$ |
| $k = 10$ | A1 | Final answer $t=10$ is A0 |
| | **4** | |
### Part 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Max speed when $2t^{-\frac{1}{2}} - \frac{3}{5}t^{\frac{1}{2}} = 0$ | M1 | Attempt to solve $a=0$ and find a value of $t$ |
| $t = \frac{10}{3}$ | A1 | |
| Maximum speed $= 4.87\text{ ms}^{-1}$ to 3 sf | B1 | Allow maximum speed as $\frac{8\sqrt{30}}{9}$ |
| | **3** | |
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt at integration of their $v$ | \*M1 | Power of at least one term increased by 1 and coefficient changed |
| $s = \frac{8}{3}t^{\frac{3}{2}} - \frac{4}{25}t^{\frac{5}{2}} [+C]$ | A1 | Correct $s$ |
| Substitute their $t = \frac{10}{3}$ and $t = 10$ | DM1 | Use their $t = 10$ and their $t = \frac{10}{3} (\neq 0)$ correctly |
| Distance $= 20.7$ m | A1 | Distance $= 20.7479...$ |
---6 A cyclist starts from rest at a fixed point $O$ and moves in a straight line, before coming to rest $k$ seconds later. The acceleration of the cyclist at time $t \mathrm {~s}$ after leaving $O$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 2 t ^ { - \frac { 1 } { 2 } } - \frac { 3 } { 5 } t ^ { \frac { 1 } { 2 } }$ for $0 < t \leqslant k$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the maximum speed of the cyclist.
\item Find an expression for the displacement from $O$ in terms of $t$. Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q6 [11]}}