| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Standard +0.3 This question requires differentiation to find acceleration, solving a quadratic equation, and integration to find distance. While it involves multiple steps and non-constant acceleration, the techniques are straightforward applications of standard calculus methods with no conceptual surprises—slightly above average due to the multi-part nature and need to recognize when direction changes (v=0). |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([a = 1.5t - 0.1875t^2]\) | M1 | For using \(a = dv/dt\) |
| \([0.1875t(8-t) = 0]\) | DM1 | For attempting to solve \(dv/dt = 0\) |
| Acceleration is zero when \(t = 8\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Changes direction when \(t = 12\) | B1 | |
| M1 | For using \(s = \int v\,dt\) | |
| \(s = 0.25t^3 - 0.0625t^4 \div 4 \quad (+C)\) | A1 | |
| \([s = 0.25 \times 1728 - 0.0625 \times 20736 \div 4]\) | DM1 | For using limits 0 to (12) or equivalent |
| Distance is 108 m | A1 [5] |
## Question 4(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[a = 1.5t - 0.1875t^2]$ | M1 | For using $a = dv/dt$ |
| $[0.1875t(8-t) = 0]$ | DM1 | For attempting to solve $dv/dt = 0$ |
| Acceleration is zero when $t = 8$ | A1 [3] | |
## Question 4(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Changes direction when $t = 12$ | B1 | |
| | M1 | For using $s = \int v\,dt$ |
| $s = 0.25t^3 - 0.0625t^4 \div 4 \quad (+C)$ | A1 | |
| $[s = 0.25 \times 1728 - 0.0625 \times 20736 \div 4]$ | DM1 | For using limits 0 to (12) or equivalent |
| Distance is 108 m | A1 [5] | |
---4 A particle $P$ starts at the point $O$ and travels in a straight line. At time $t$ seconds after leaving $O$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $v = 0.75 t ^ { 2 } - 0.0625 t ^ { 3 }$. Find\\
(i) the positive value of $t$ for which the acceleration is zero,\\
(ii) the distance travelled by $P$ before it changes its direction of motion.
\hfill \mbox{\textit{CAIE M1 2012 Q4 [8]}}