| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration question requiring integration and differentiation of standard functions (powers of t). Part (i) involves setting a=0 and solving a simple equation. Part (ii) requires finding when da/dt=0, then integrating to find velocity at that time. All techniques are routine M1 procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | [1.2T1/2 – 0.6T = 0] | M1 |
| T1/2 = 2 → T = 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | [da/dt = 0.6t1/2 – 0.6] | M1 |
| t = 1 | A1 | Solve da/dt = 0 and find t |
| [v = 0.8t 3/2 – 0.3t2 (+ C)] | M1 | Attempt to integrate a to find v |
| A1 | Correct integration | |
| [C = 1] | M1 | Use v = 1 at t = 0 either finding C or by using limits as |
| Answer | Marks | Guidance |
|---|---|---|
| Velocity when acceleration is max is 1.5 ms-1 | A1 | v = 1.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | [1.2T1/2 – 0.6T = 0] | M1 | Attempt to find time of maximum v, set a = 0 and solve for T
T1/2 = 2 → T = 4 | A1
2
--- 5(ii) ---
5(ii) | [da/dt = 0.6t1/2 – 0.6] | M1 | Attempt to differentiate a
t = 1 | A1 | Solve da/dt = 0 and find t
[v = 0.8t 3/2 – 0.3t2 (+ C)] | M1 | Attempt to integrate a to find v
A1 | Correct integration
[C = 1] | M1 | Use v = 1 at t = 0 either finding C or by using limits as
v(1) – v(0) = [0.8(1)3/2 – 0.3(1)2] – [0.8(0)3/2 – 0.3(0)2]
Velocity when acceleration is max is 1.5 ms-1 | A1 | v = 1.5
6
Question | Answer | Marks | GuidanceA particle moves in a straight line starting from a point $O$ with initial velocity $1$ m s$^{-1}$. The acceleration of the particle at time $t$ s after leaving $O$ is $a$ m s$^{-2}$, where
$$a = 1.2t^{\frac{1}{2}} - 0.6t.$$
\begin{enumerate}[label=(\roman*)]
\item At time $T$ s after leaving $O$ the particle reaches its maximum velocity. Find the value of $T$. [2]
\item Find the velocity of the particle when its acceleration is maximum (you do not need to verify that the acceleration is a maximum rather than a minimum). [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q5 [8]}}