| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration problem requiring integration with initial conditions. Students must integrate twice (a→v→s) and apply the given condition v=0 at t=20 to find the constant. Finding the second rest time involves solving a quadratic, and the distance calculation requires evaluating a definite integral. All steps are standard M1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Attempt to integrate \(a\) | |
| \(v = 6t - 0.12t^2\ (+c)\) | A1 | |
| \(0 = 6\times 20 - 0.12\times 20^2 + c\) | DM1 | Substitute \(v=0\), \(t=20\) in an equation with arbitrary constant |
| \(0.12t^2 - 6t + 72 = 0\) | DM1 | Substitute \(v=0\) and attempt to solve a 3-term quadratic |
| \(t = 30\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s = 3t^2 - 0.04t^3 - 72t\ (+k)\) | M1 | Attempt to integrate \(v\) |
| \(s(30) - s(20) = -540 - (-560)\) | DM1 | Use of limits 20 and then 30 |
| Distance travelled \(= 20\) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Attempt to integrate $a$ |
| $v = 6t - 0.12t^2\ (+c)$ | A1 | |
| $0 = 6\times 20 - 0.12\times 20^2 + c$ | DM1 | Substitute $v=0$, $t=20$ in an equation with arbitrary constant |
| $0.12t^2 - 6t + 72 = 0$ | DM1 | Substitute $v=0$ and attempt to solve a 3-term quadratic |
| $t = 30$ | A1 | |
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## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s = 3t^2 - 0.04t^3 - 72t\ (+k)$ | M1 | Attempt to integrate $v$ |
| $s(30) - s(20) = -540 - (-560)$ | DM1 | Use of limits 20 and then 30 |
| Distance travelled $= 20$ | A1 | |6 A particle $P$ moves in a straight line passing through a point $O$. At time $t \mathrm {~s}$, the acceleration, $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, of $P$ is given by $a = 6 - 0.24 t$. The particle comes to instantaneous rest at time $t = 20$.\\
(i) Find the value of $t$ at which the particle is again at instantaneous rest.\\
(ii) Find the distance the particle travels between the times of instantaneous rest.\\
\hfill \mbox{\textit{CAIE M1 2018 Q6 [8]}}