| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Sketching velocity-time graphs |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: substituting values to verify v=0, differentiating to find acceleration=0, sketching a cubic velocity graph, and integrating to find displacement. All steps are routine A-level mechanics procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks |
|---|---|
| 7 (i) | [0.0001t(t – 50)(t – 100) = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| v(t) = 0 when t = 0, 50 & 100 | M1 | |
| A1 | 2 | Either factorise v(t) and solve v(t) = 0 |
| Answer | Marks |
|---|---|
| (ii) | [0.0003t 2 – 0.03t + 0.5 = 0] |
| Answer | Marks |
|---|---|
| 2 | M1 |
| dM1 | dv |
| Answer | Marks | Guidance |
|---|---|---|
| Page 8 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 43 |
| Answer | Marks |
|---|---|
| the point (50, 0). | A1 |
| Answer | Marks |
|---|---|
| B1 | 7 |
| (iii) | 4 3 2 |
| Answer | Marks |
|---|---|
| Greatest distance is 156 m | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | For integrating v(t) to obtain s(t) |
Question 7:
--- 7 (i) ---
7 (i) | [0.0001t(t – 50)(t – 100) = 0
or v(0) = 0, v(50) = 0, v(100) = 0]
v(t) = 0 when t = 0, 50 & 100 | M1
A1 | 2 | Either factorise v(t) and solve v(t) = 0
or evaluate v(0) , v(50) and v(100)
(ii) | [0.0003t 2 – 0.03t + 0.5 = 0]
t 2 – 100t + 1667 = 0 (cid:1)
1 ( )
t = 100± 1002 −4×1667
2 | M1
dM1 | dv
For using a(t) =
dt
For solving a(t) = 0
Page 8 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 43
a = 0 when t = 21.1 and when t = 78.9
v(21.1) = 4.81
v(78.9) = –4.81
Convex curve from (0,0) to (50,0) with
v > 0 and has a maximum point.
The curve for (50, 0) to (100, 0) is exactly
the same as the first curve positioned by
°
rotating the first curve through 180 about
the point (50, 0). | A1
B1
B1
B1
B1 | 7
(iii) | 4 3 2
s(t) = 0.000025t – 0.005t + 0.25t (+ c)
[156.25 – 625 + 625]
Greatest distance is 156 m | M1
A1
M1
A1 | 4 | For integrating v(t) to obtain s(t)
For using lower and upper limits of 0 and
50 respectively.A particle $P$ moves on a straight line. It starts at a point $O$ on the line and returns to $O$ 100 s later. The velocity of $P$ is $v \text{ m s}^{-1}$ at time $t$ s after leaving $O$, where
$$v = 0.0001t^3 - 0.015t^2 + 0.5t.$$
\begin{enumerate}[label=(\roman*)]
\item Show that $P$ is instantaneously at rest when $t = 0$, $t = 50$ and $t = 100$. [2]
\item Find the values of $v$ at the times for which the acceleration of $P$ is zero, and sketch the velocity-time graph for $P$'s motion for $0 \leq t \leq 100$. [7]
\item Find the greatest distance of $P$ from $O$ for $0 \leq t \leq 100$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2015 Q7 [13]}}