| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from displacement differentiation |
| Difficulty | Moderate -0.8 This is a straightforward calculus application question requiring differentiation of a polynomial to find velocity, solving a quadratic equation for when v=0, and finding a minimum using standard optimization techniques (differentiate v, set equal to zero). All three parts are routine A-level mechanics procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | M1 | Attempt differentiation |
| v = 3t 2 – 8t + 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | 3t 2 – 8t + 4 = 0 | M1 |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iii) | [6t – 8 = 0] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 3 | M1 | Solve for t and attempt v |
| Answer | Marks |
|---|---|
| 3 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 | M1 | Attempt to complete the square for v |
| Answer | Marks | Guidance |
|---|---|---|
| 3 3 3 | M1 | Find value of t for minimum v and attempt |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | M1 | Attempt differentiation
v = 3t 2 – 8t + 4 | A1
2
--- 4(ii) ---
4(ii) | 3t 2 – 8t + 4 = 0 | M1 | Set v = 0 and attempt to solve a relevant 3
term quadratic
2
t = and t = 2
3 | A1
2
Question | Answer | Marks | Guidance
--- 4(iii) ---
4(iii) | [6t – 8 = 0] | M1 | Differentiate v and equate to 0
4 4 4
[t = , v =3( )2 – 8( ) + 4]
3 3 3 | M1 | Solve for t and attempt v
4
v = –
3 | A1
3
Alternative scheme for Question 4(iii)
8 4
[v = 3(t 2 – t) + 4 = 3(t – )2 +......]
3 3 | M1 | Attempt to complete the square for v
4 4 4
[t = , v = 3(t – )2 – ]
3 3 3 | M1 | Find value of t for minimum v and attempt
to find v
4
v = –
3 | A1
Question | Answer | Marks | GuidanceA particle $P$ moves in a straight line starting from a point $O$. At time $t \text{ s}$ after leaving $O$, the displacement $s \text{ m}$ from $O$ is given by $s = t^3 - 4t^2 + 4t$ and the velocity is $v \text{ m s}^{-1}$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $v$ in terms of $t$. [2]
\item Find the two values of $t$ for which $P$ is at instantaneous rest. [2]
\item Find the minimum velocity of $P$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q4 [7]}}