| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a standard M1 piecewise kinematics problem requiring integration of acceleration to find velocity and distance, plus basic calculus (differentiation to find maximum). While it involves multiple steps and careful handling of piecewise functions with continuity conditions, the techniques are routine for M1 students: differentiate to find maximum, integrate twice with initial conditions, and solve for when velocity equals zero. No novel insight required. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([15 - 6t = 0]\) | M1 | For differentiation |
| Max acceleration when \(t = 2.5\text{ s}\) | A1 | May be stated from an \(a\)-\(t\) diagram |
| Max acceleration \(= 18.75\text{ ms}^{-2}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Speed \(= 7.5t^2 - t^3\ (+c)\) | M1 | For using integration to obtain speed |
| Distance \(= 2.5t^3 - 0.25t^4\ (+ct+d)\) | M1 | For using integration to obtain distance |
| \(= 2.5\times125 - 0.25\times625 = 156.25\text{ m}\) | A1 | [3] Allow distance \(= 625/4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v(5) = 7.5 \times 25 - 125 = 62.5 \text{ ms}^{-1}\) | B1 | Allow \(v(5) = 125/2\) |
| \(\int_{5}^{k} -\frac{625}{t^2}\, dt = \left[\frac{625}{t}\right]_{5}^{k}\) | M1 | Integral with correct limits |
| \(= \frac{625}{k} - \frac{625}{5} = \frac{625}{k} - 125\) | A1 | |
| \(\frac{625}{k} - 125 = v(k) - v(5) = -62.5\) | M1 | Use of \(v(5) = 62.5\) and \(v(k) = 0\) |
| \(k = 10\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v(5) = 7.5 \times 25 - 125 = 62.5 \text{ ms}^{-1}\) | B1 | |
| \(v(t) = \int -\frac{625}{t^2}\, dt = \frac{625}{t} + c\) | M1 | Using indefinite integration |
| \([c = -62.5]\) | For using \(v(5) = 62.5\) to find \(c\) | |
| \(v(t) = \frac{625}{t} - 62.5\) | A1 | and setting \(v(k) = 0\) |
| \(v(k) = \frac{625}{k} - 62.5 = 0\) | M1 | |
| \(k = 10\) | A1 | [5] |
## Question 7(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[15 - 6t = 0]$ | M1 | For differentiation |
| Max acceleration when $t = 2.5\text{ s}$ | A1 | May be stated from an $a$-$t$ diagram |
| Max acceleration $= 18.75\text{ ms}^{-2}$ | A1 | [3] |
## Question 7(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Speed $= 7.5t^2 - t^3\ (+c)$ | M1 | For using integration to obtain speed |
| Distance $= 2.5t^3 - 0.25t^4\ (+ct+d)$ | M1 | For using integration to obtain distance |
| $= 2.5\times125 - 0.25\times625 = 156.25\text{ m}$ | A1 | [3] Allow distance $= 625/4$ |
## Question 7(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v(5) = 7.5 \times 25 - 125 = 62.5 \text{ ms}^{-1}$ | **B1** | Allow $v(5) = 125/2$ |
| $\int_{5}^{k} -\frac{625}{t^2}\, dt = \left[\frac{625}{t}\right]_{5}^{k}$ | **M1** | Integral with correct limits |
| $= \frac{625}{k} - \frac{625}{5} = \frac{625}{k} - 125$ | **A1** | |
| $\frac{625}{k} - 125 = v(k) - v(5) = -62.5$ | **M1** | Use of $v(5) = 62.5$ and $v(k) = 0$ |
| $k = 10$ | **A1** | [5] |
**Alternative for 7(iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v(5) = 7.5 \times 25 - 125 = 62.5 \text{ ms}^{-1}$ | **B1** | |
| $v(t) = \int -\frac{625}{t^2}\, dt = \frac{625}{t} + c$ | **M1** | Using indefinite integration |
| $[c = -62.5]$ | | For using $v(5) = 62.5$ to find $c$ |
| $v(t) = \frac{625}{t} - 62.5$ | **A1** | and setting $v(k) = 0$ |
| $v(k) = \frac{625}{k} - 62.5 = 0$ | **M1** | |
| $k = 10$ | **A1** | [5] |7 A racing car is moving in a straight line. The acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ at time $t \mathrm {~s}$ after the car starts from rest is given by
$$\begin{array} { l l }
a = 15 t - 3 t ^ { 2 } & \text { for } 0 \leqslant t \leqslant 5 \\
a = - \frac { 625 } { t ^ { 2 } } & \text { for } 5 < t \leqslant k
\end{array}$$
where $k$ is a constant.\\
(i) Find the maximum acceleration of the car in the first five seconds of its motion.\\
(ii) Find the distance of the car from its starting point when $t = 5$.\\
(iii) The car comes to rest when $t = k$. Find the value of $k$.
\hfill \mbox{\textit{CAIE M1 2016 Q7 [11]}}