| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question requiring integration of velocity to find displacement (with a constant of integration determined from initial conditions) and then solving a cubic equation. Both parts use standard M1 techniques with no novel problem-solving required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 5 | (i) |
| (ii) | 3 |
| Answer | Marks |
|---|---|
| Velocity is 3ms | M1 |
| Answer | Marks |
|---|---|
| B1ft 3 | For attempting to use |
Question 5:
5 | (i)
(ii) | 3
x = 0.01t (+C)
× 3
2.5 = 0.01 5 + C
3
x = 0.01t + 1.25
3
0.01t + 1.25 = 11.25
t = 10
-1
Velocity is 3ms | M1
A1
DM1
A1 ft 4
M1
A1
B1ft 3 | For attempting to use
x(t) = ∫vdt
For substituting x = 2.5 and
t = 5 and attempting to find C
ft candidate’s a where
3
x = at + C
For attempting to solve
x(t) = 11.25 (equation needs to
3
be of the form at = b)
2
ft for value of 0.03tA particle $P$ moves along the $x$-axis in the positive direction. The velocity of $P$ at time $t \text{ s}$ is $0.03t^2 \text{ m s}^{-1}$. When $t = 5$ the displacement of $P$ from the origin $O$ is $2.5 \text{ m}$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression, in terms of $t$, for the displacement of $P$ from $O$.
[4]
\item Find the velocity of $P$ when its displacement from $O$ is $11.25 \text{ m}$.
[3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2005 Q5 [7]}}