| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from displacement differentiation |
| Difficulty | Moderate -0.8 This is a straightforward differentiation exercise requiring only basic calculus rules (power rule applied twice) and simple algebraic manipulation. The question tests routine application of v = dx/dt and a = dv/dt with no problem-solving insight needed, making it easier than average for A-level mechanics. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07i Differentiate x^n: for rational n and sums3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks |
|---|---|
| For differentiating \(x\) | M1 |
| \(\dot{x} = t + \frac{1}{10}t^2\) | A1 |
| Speed is 20 ms\(^{-1}\) | A1 |
| Answer | Marks |
|---|---|
| \(\ddot{x} = 1 + \frac{1}{5}t\) | B1ft |
| For attempting to solve \(\ddot{x}(t) = 2\ddot{x}(0)\) \(\left(1 + \frac{1}{5}t = 2\right)\) | M1 |
| \(t = 5\) | A1 |
# Question 4:
## Part (i)
| For differentiating $x$ | M1 | |
| $\dot{x} = t + \frac{1}{10}t^2$ | A1 | |
| Speed is 20 ms$^{-1}$ | A1 | |
## Part (ii)
| $\ddot{x} = 1 + \frac{1}{5}t$ | B1ft | |
| For attempting to solve $\ddot{x}(t) = 2\ddot{x}(0)$ $\left(1 + \frac{1}{5}t = 2\right)$ | M1 | |
| $t = 5$ | A1 | |
---4 A particle moves in a straight line. Its displacement $t$ seconds after leaving the fixed point $O$ is $x$ metres, where $x = \frac { 1 } { 2 } t ^ { 2 } + \frac { 1 } { 30 } t ^ { 3 }$. Find\\
(i) the speed of the particle when $t = 10$,\\
(ii) the value of $t$ for which the acceleration of the particle is twice its initial acceleration.
\hfill \mbox{\textit{CAIE M1 2003 Q4 [6]}}