| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise kinematics problem requiring integration of given velocity functions and substitution. Part (i) is routine integration, part (ii) requires finding a constant of integration to ensure continuity, and part (iii) involves solving a simple equation. All techniques are standard M1 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = 4t^2 - \frac{2t^3}{3}\ (+C)\) | M1, A1 | For using \(s = \int v\, dt\) |
| \(s = 4 \times 9 - \frac{2}{3} \times 27\) | M1 | For using limits 0, 3 or equivalent (may be implied by absence of \(C\) and substituting \(t=3\)) |
| Distance is \(18\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = -\frac{54}{t}\ (+C)\) | B1 | |
| \(18 = -\frac{54}{3} + C\) | M1 | For using \(s(3) = \text{ans(i)}\) |
| Displacement is \(36 - \frac{54}{t}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(36 - \frac{54}{t} = 27 \Rightarrow t = 6\) | M1* | For solving \(s(t) = 27\) |
| \(v = \frac{54}{36}\) | M1*dep | For substituting value of \(t\) found into \(v(t)\) |
| \(v = 1.5\) | A1 |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 4t^2 - \frac{2t^3}{3}\ (+C)$ | M1, A1 | For using $s = \int v\, dt$ |
| $s = 4 \times 9 - \frac{2}{3} \times 27$ | M1 | For using limits 0, 3 or equivalent (may be implied by absence of $C$ and substituting $t=3$) |
| Distance is $18$ m | A1 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = -\frac{54}{t}\ (+C)$ | B1 | |
| $18 = -\frac{54}{3} + C$ | M1 | For using $s(3) = \text{ans(i)}$ |
| Displacement is $36 - \frac{54}{t}$ | A1 | |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $36 - \frac{54}{t} = 27 \Rightarrow t = 6$ | M1* | For solving $s(t) = 27$ |
| $v = \frac{54}{36}$ | M1*dep | For substituting value of $t$ found into $v(t)$ |
| $v = 1.5$ | A1 | |
---6 A particle $P$ starts from rest at $O$ and travels in a straight line. Its velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ is given by $v = 8 t - 2 t ^ { 2 }$ for $0 \leqslant t \leqslant 3$, and $v = \frac { 54 } { t ^ { 2 } }$ for $t > 3$. Find\\
(i) the distance travelled by $P$ in the first 3 seconds,\\
(ii) an expression in terms of $t$ for the displacement of $P$ from $O$, valid for $t > 3$,\\
(iii) the value of $v$ when the displacement of $P$ from $O$ is 27 m .
\hfill \mbox{\textit{CAIE M1 2005 Q6 [10]}}