| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.8 This is a straightforward velocity-time graph question requiring basic kinematics: using area under graph for displacement and gradient for acceleration. All parts involve direct application of standard formulas (v=u+at, area of trapezium) with no problem-solving insight needed. Easier than average A-level due to step-by-step scaffolding and routine calculations. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| 6 | (i) |
| Answer | Marks |
|---|---|
| (iv) | ± |
| Answer | Marks |
|---|---|
| Deceleration is 4/3 ms | M1 |
| Answer | Marks |
|---|---|
| A1ft 2 | For using the idea that the area |
| Answer | Marks | Guidance |
|---|---|---|
| Page 4 | Mark Scheme | Syllabus |
| A AND AS LEVEL – JUNE 2005 | 9709 | 4 |
Question 6:
6 | (i)
(ii)
(iii)
(iv) | ±
½ 5v = 10
max
-1
Greatest speed is 4 ms
×
V/3 = 2 or V = 0 + 2 3
V = 6
½ (T + 9.5)6 = 34.5 or
½ (t – 18 + 9.5)6 = 34.5
Time is 2 s
6
d =
24.5 − (18 + 2)
-2
Deceleration is 4/3 ms | M1
A1 2
M1
A1 2
M1
A1 ft
A1 3
M1
A1ft 2 | For using the idea that the area
of the relevant triangle
represents distance
For using the idea that the
gradient represents acceleration
or v = 0 + at
For an attempt to find the area
of the trapezium in terms of T
(or of t) and equate with 34.5
Any correct form of equation in
T (ot t)
For using the idea that minus
the gradient represents
deceleration
Page 4 | Mark Scheme | Syllabus | Paper
A AND AS LEVEL – JUNE 2005 | 9709 | 4\includegraphics{figure_6}
The diagram shows the velocity-time graph for a lift moving between floors in a building. The graph consists of straight line segments. In the first stage the lift travels downwards from the ground floor for $5 \text{ s}$, coming to rest at the basement after travelling $10 \text{ m}$.
\begin{enumerate}[label=(\roman*)]
\item Find the greatest speed reached during this stage.
[2]
\end{enumerate}
The second stage consists of a $10 \text{ s}$ wait at the basement. In the third stage, the lift travels upwards until it comes to rest at a floor $34.5 \text{ m}$ above the basement, arriving $24.5 \text{ s}$ after the start of the first stage. The lift accelerates at $2 \text{ m s}^{-2}$ for the first $3 \text{ s}$ of the third stage, reaching a speed of $V \text{ m s}^{-1}$. Find
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the value of $V$,
[2]
\item the time during the third stage for which the lift is moving at constant speed,
[3]
\item the deceleration of the lift in the final part of the third stage.
[2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2005 Q6 [9]}}