| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question requiring integration of acceleration to find velocity, then solving a quadratic equation. The techniques are standard M2 content with clear initial conditions and no conceptual complications, making it slightly easier than average but still requiring correct application of calculus and algebraic manipulation. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| \(v = \int a \, dt = 6t^2 - 15t + c\) | M1 A1 A1 |
| \(v(0) = 0\); \(c = 0\) | |
| \(v = 6t^2 - 15t\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(6t^2 - 15t - 36 = 0\) | M1 A1 A1 | |
| \(3(2t + 3)(t - 4) = 0\) | ||
| \(t = 4\) | Total: 6 marks |
### (a)
$v = \int a \, dt = 6t^2 - 15t + c$ | M1 A1 A1 |
$v(0) = 0$; $c = 0$ | |
$v = 6t^2 - 15t$ | |
### (b)
$6t^2 - 15t - 36 = 0$ | M1 A1 A1 |
$3(2t + 3)(t - 4) = 0$ | |
$t = 4$ | | **Total: 6 marks**A particle $P$, initially at rest at the point $O$, moves in a straight line such that at time $t$ seconds after leaving $O$ its acceleration is $(12t - 15)$ ms$^{-2}$. Find
\begin{enumerate}[label=(\alph*)]
\item the velocity of $P$ at time $t$ seconds after it leaves $O$, [3 marks]
\item the value of $t$ when the speed of $P$ is 36 ms$^{-1}$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [6]}}