| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring integration of velocity to find displacement, substitution of boundary conditions to form simultaneous equations, and differentiation to find minimum acceleration. While it involves multiple steps and algebraic manipulation, the techniques are routine for M1 level with no novel problem-solving insight required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(10 = 0.04\times5^3 + 5^2c + 5k\) \((5c+k=1)\) | B1 | Use of \(t=5\), \(v=10\) |
| \(s = \frac{0.04}{4}t^4 + \frac{ct^3}{3} + \frac{kt^2}{2} + (C)\) | *M1 | For use of \(s = \int v\,dt\) |
| \(25 = 0.01\times5^4 + \frac{5^3}{3}c + \frac{5^2}{2}k\) | DM1 | Use of \(t=0\), \(s=0\) and \(t=5\), \(s=25\) |
| \(6.25 + \frac{125}{3}c + \frac{25}{2}k = 25\) \(\left(\frac{125}{3}c + \frac{25}{2}k = 18.75\right)\) | A1 | |
| Solving for \(c\) or for \(k\) | M1 | |
| \(c = -0.3\) and \(k = 2.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 0.12t^2 - 0.6t + 2.5\) | M1 | For use of \(a = \frac{dv}{dt}\) |
| \(a' = 0.24t - 0.6 = 0 \rightarrow t = \ldots\) or \(a = 0.12(t^2-5t+\ldots) = 0.12\left[(t-2.5)^2+\ldots\right]\) | M1 | Uses \(\frac{da}{dt}=0\) or completes the square for \(a\) |
| Minimum when \(t = 2.5\) | A1 | AG |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $10 = 0.04\times5^3 + 5^2c + 5k$ $(5c+k=1)$ | B1 | Use of $t=5$, $v=10$ |
| $s = \frac{0.04}{4}t^4 + \frac{ct^3}{3} + \frac{kt^2}{2} + (C)$ | *M1 | For use of $s = \int v\,dt$ |
| $25 = 0.01\times5^4 + \frac{5^3}{3}c + \frac{5^2}{2}k$ | DM1 | Use of $t=0$, $s=0$ and $t=5$, $s=25$ |
| $6.25 + \frac{125}{3}c + \frac{25}{2}k = 25$ $\left(\frac{125}{3}c + \frac{25}{2}k = 18.75\right)$ | A1 | |
| Solving for $c$ or for $k$ | M1 | |
| $c = -0.3$ and $k = 2.5$ | A1 | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 0.12t^2 - 0.6t + 2.5$ | M1 | For use of $a = \frac{dv}{dt}$ |
| $a' = 0.24t - 0.6 = 0 \rightarrow t = \ldots$ **or** $a = 0.12(t^2-5t+\ldots) = 0.12\left[(t-2.5)^2+\ldots\right]$ | M1 | Uses $\frac{da}{dt}=0$ or completes the square for $a$ |
| Minimum when $t = 2.5$ | A1 | AG |6 Particle $P$ travels in a straight line from $A$ to $B$. The velocity of $P$ at time $t \mathrm {~s}$ after leaving $A$ is denoted by $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where
$$v = 0.04 t ^ { 3 } + c t ^ { 2 } + k t$$
$P$ takes 5 s to travel from $A$ to $B$ and it reaches $B$ with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The distance $A B$ is 25 m .\\
(i) Find the values of the constants $c$ and $k$.\\
(ii) Show that the acceleration of $P$ is a minimum when $t = 2.5$.\\
\hfill \mbox{\textit{CAIE M1 2019 Q6 [9]}}