| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from displacement differentiation |
| Difficulty | Easy -1.2 This is a straightforward differentiation question requiring only the application of standard rules (power rule and chain rule for cos). It's a single-step problem with no problem-solving or conceptual challenge—purely routine calculus that any M2 student should execute mechanically. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = \frac{ds}{dt}\) | M1 | Differentiating \(s\) with respect to \(t\) |
| \(v = 10t - 12\sin 4t\) | A1 | For \(10t\) |
| A1 | For \(-12\sin 4t\) |
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = \frac{ds}{dt}$ | M1 | Differentiating $s$ with respect to $t$ |
| $v = 10t - 12\sin 4t$ | A1 | For $10t$ |
| | A1 | For $-12\sin 4t$ |
---1 A particle moves along a straight line through the origin. At time $t$, the displacement, $s$, of the particle from the origin is given by
$$s = 5 t ^ { 2 } + 3 \cos 4 t$$
Find the velocity of the particle at time $t$.
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\hfill \mbox{\textit{AQA M2 2010 Q1 [3]}}