| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring integration of velocity to find displacement (part a) and differentiation to find minimum velocity (part b). The calculus operations are routine (integrating/differentiating polynomial and fractional power terms), and part (b) explicitly tells students to 'show' a given answer, removing the challenge of finding it independently. Slightly above average difficulty only due to the fractional power term. |
| 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 | Marks | Guidance |
| \(\left[s =\right]\int\left(t^2 - 8t^{\frac{3}{2}} + 10t\right)dt\) | *M1 | For attempting to integrate \(v\) |
| \(\left[s =\right]\frac{1}{3}t^3 - \frac{16}{5}t^{\frac{5}{2}} + 5t^2\ [+C]\) | A1 | Allow unsimplified |
| For correct use of correct limits | DM1 | Use of limit at \(t=0\) may be implied |
| Displacement \(= 2.13\) m (3sf) | A1 | Allow displacement \(= \frac{32}{15}\) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For attempting to differentiate \(v\) | *M1 | |
| \(\left[a =\right] 2t - 12t^{\frac{1}{2}} + 10\) | A1 | Allow unsimplified |
| \(a = 0 \Rightarrow 2t - 12t^{\frac{1}{2}} + 10 = 0\) | DM1 | Dependent on *M1. Set \(a=0\) and attempt to solve their 3 term equation in \(\sqrt{t}\) or \(t\) or \(p\ (=\sqrt{t})\) by treating it as a quadratic equation |
| \(2\left(t^{\frac{1}{2}}-5\right)\left(t^{\frac{1}{2}}-1\right)=0\) leading to \(t=1\) or \(t=25\) | A1 | Both correct |
| \(\frac{da}{dt} = 2 - 6t^{-\frac{1}{2}}\) | *DM1 | Dependent on *M1. Determine the nature of the stationary point by differentiating \(a\) and testing the sign of \(\frac{da}{dt}\), or by substituting values either side of their \(t\) value(s). If using \(\frac{da}{dt}\) then must evaluate it at a \(t\) value for M1. Allow use with any \(t\) value from *their* 'quadratic' |
| Use \(t=25\) in \(\frac{da}{dt} = 2 - 6\times 25^{-\frac{1}{2}}\); Evaluating \(\frac{da}{dt}\) correctly, hence a minimum | A1 | Or by using a convincing argument to show that \(t=25\) gives a minimum value of \(v\). If evaluated then \(\frac{da}{dt}\) must be \(0.8\) |
| Minimum velocity \(= 25^2 - 8\times 25^{\frac{3}{2}} + 10\times 25 = -125 \text{ ms}^{-1}\) | B1 | AG This mark is awarded only if the previous 6 marks are awarded |
| 7 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[s =\right]\int\left(t^2 - 8t^{\frac{3}{2}} + 10t\right)dt$ | *M1 | For attempting to integrate $v$ |
| $\left[s =\right]\frac{1}{3}t^3 - \frac{16}{5}t^{\frac{5}{2}} + 5t^2\ [+C]$ | A1 | Allow unsimplified |
| For correct use of correct limits | DM1 | Use of limit at $t=0$ may be implied |
| Displacement $= 2.13$ m (3sf) | A1 | Allow displacement $= \frac{32}{15}$ |
| | **4** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For attempting to differentiate $v$ | *M1 | |
| $\left[a =\right] 2t - 12t^{\frac{1}{2}} + 10$ | A1 | Allow unsimplified |
| $a = 0 \Rightarrow 2t - 12t^{\frac{1}{2}} + 10 = 0$ | DM1 | Dependent on *M1. Set $a=0$ and attempt to solve their 3 term equation in $\sqrt{t}$ or $t$ or $p\ (=\sqrt{t})$ by treating it as a quadratic equation |
| $2\left(t^{\frac{1}{2}}-5\right)\left(t^{\frac{1}{2}}-1\right)=0$ leading to $t=1$ or $t=25$ | A1 | Both correct |
| $\frac{da}{dt} = 2 - 6t^{-\frac{1}{2}}$ | *DM1 | Dependent on *M1. Determine the nature of the stationary point by differentiating $a$ and testing the sign of $\frac{da}{dt}$, or by substituting values either side of their $t$ value(s). If using $\frac{da}{dt}$ then must evaluate it at a $t$ value for M1. Allow use with any $t$ value from *their* 'quadratic' |
| Use $t=25$ in $\frac{da}{dt} = 2 - 6\times 25^{-\frac{1}{2}}$; Evaluating $\frac{da}{dt}$ correctly, hence a minimum | A1 | Or by using a convincing argument to show that $t=25$ gives a minimum value of $v$. If evaluated then $\frac{da}{dt}$ must be $0.8$ |
| Minimum velocity $= 25^2 - 8\times 25^{\frac{3}{2}} + 10\times 25 = -125 \text{ ms}^{-1}$ | B1 | AG This mark is awarded only if the previous 6 marks are awarded |
| | **7** | |6 A particle moves in a straight line. It starts from rest from a fixed point $O$ on the line. Its velocity at time $t \mathrm {~s}$ after leaving $O$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = t ^ { 2 } - 8 t ^ { \frac { 3 } { 2 } } + 10 t$.
\begin{enumerate}[label=(\alph*)]
\item Find the displacement of the particle from $O$ when $t = 1$.
\item Show that the minimum velocity of the particle is $- 125 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q6 [11]}}