| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.8 This question requires integration of a linear acceleration function, evaluation at specific times, then solving a piecewise constant acceleration problem for particle B and comparing displacements. It involves multiple steps across different motion phases, requiring careful bookkeeping of velocities and positions, but uses standard kinematics techniques without requiring novel insight. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For integrating \(a(t)\) to find \(v(t)\) | |
| \(v(t) = 0.05t - 0.0001t^2\ (+0)\) | A1 | |
| \(v(200) = 10 - 4 = 6\) ms\(^{-1}\) | A1 | |
| \(v(500) = 25 - 25 = 0\) | A1 | |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For integrating \(v(t)\) between limits 0 to 500 to obtain the distance \(A\) travels | |
| \(\int_0^{500}(0.05t - 0.0001t^2)\,dt\) | ||
| \(\left[\frac{0.05t^2}{2} - \frac{0.0001t^3}{3}\right]_0^{500}\) | A1 | |
| Distance \(= 0.025 \times 500^2 - 0.0001 \times 500^3 \div 3 = 2083\) m | A1 | Accept 2080 |
| M1 | For using area property of graph or \(s = \frac{1}{2}(u+v)t\) or \(s = ut + \frac{1}{2}at^2\) to find distance travelled by \(B\) | |
| Distance \(= \frac{1}{2} \times 6 \times 500 = 1500\) m | A1 | |
| or distance \(= \frac{1}{2}(0+6)\times200 + \frac{1}{2}(6+0)\times300\) | ||
| or distance \(= \left(0 + \frac{1}{2}0.03\times200^2\right) + \left(6\times300 + \frac{1}{2}(-0.02)300^2\right)\) | A1 | |
| Distance between \(A\) and \(B\) is \(2083 - 1500 = 583\) m | B1\(\checkmark\) | Can only be scored if distance travelled by \(A\) has been found using integration |
| Total: 6 marks |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For integrating $a(t)$ to find $v(t)$ |
| $v(t) = 0.05t - 0.0001t^2\ (+0)$ | A1 | |
| $v(200) = 10 - 4 = 6$ ms$^{-1}$ | A1 | |
| $v(500) = 25 - 25 = 0$ | A1 | |
| **Total: 4 marks** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For integrating $v(t)$ between limits 0 to 500 to obtain the distance $A$ travels |
| $\int_0^{500}(0.05t - 0.0001t^2)\,dt$ | | |
| $\left[\frac{0.05t^2}{2} - \frac{0.0001t^3}{3}\right]_0^{500}$ | A1 | |
| Distance $= 0.025 \times 500^2 - 0.0001 \times 500^3 \div 3 = 2083$ m | A1 | Accept 2080 |
| | M1 | For using area property of graph or $s = \frac{1}{2}(u+v)t$ or $s = ut + \frac{1}{2}at^2$ to find distance travelled by $B$ |
| Distance $= \frac{1}{2} \times 6 \times 500 = 1500$ m | A1 | |
| or distance $= \frac{1}{2}(0+6)\times200 + \frac{1}{2}(6+0)\times300$ | | |
| or distance $= \left(0 + \frac{1}{2}0.03\times200^2\right) + \left(6\times300 + \frac{1}{2}(-0.02)300^2\right)$ | A1 | |
| Distance between $A$ and $B$ is $2083 - 1500 = 583$ m | B1$\checkmark$ | Can only be scored if distance travelled by $A$ has been found using integration |
| **Total: 6 marks** | | |
---6 Two particles $A$ and $B$ start to move at the same instant from a point $O$. The particles move in the same direction along the same straight line. The acceleration of $A$ at time $t \mathrm {~s}$ after starting to move is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where $a = 0.05 - 0.0002 t$.\\
(i) Find A's velocity when $t = 200$ and when $t = 500$.\\
$B$ moves with constant acceleration for the first 200 s and has the same velocity as $A$ when $t = 200 . B$ moves with constant retardation from $t = 200$ to $t = 500$ and has the same velocity as $A$ when $t = 500$.\\
(ii) Find the distance between $A$ and $B$ when $t = 500$.
\hfill \mbox{\textit{CAIE M1 2015 Q6 [10]}}