| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from displacement differentiation |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring differentiation of an exponential function and solving a simple equation. Part (a) is direct substitution, and part (b) requires finding v = dx/dt = 1 - (1/10)e^t = 0, then solving e^t = 10 using logarithms. While it involves exponential functions (making it slightly beyond basic calculus), the steps are routine and well-practiced for M2 students with no problem-solving insight required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) when \(t = 0, x = 2 + 0 - \frac{1}{10} = 1.9\) m | M1 A1 | |
| (b) \(v = \frac{dx}{dt} = 1 - \frac{1}{10}e^t\) | A1 | |
| at rest when \(v = 0\): \(1 - \frac{1}{10}e^t = 0 \quad \therefore e^t = 10\) | M1 A1 | |
| \(t = \ln 10 = 2.3\) (1dp) | A1 | (6) |
**(a)** when $t = 0, x = 2 + 0 - \frac{1}{10} = 1.9$ m | M1 A1 |
**(b)** $v = \frac{dx}{dt} = 1 - \frac{1}{10}e^t$ | A1 |
at rest when $v = 0$: $1 - \frac{1}{10}e^t = 0 \quad \therefore e^t = 10$ | M1 A1 |
$t = \ln 10 = 2.3$ (1dp) | A1 | (6)2. A particle $P$ moves along the $x$-axis such that its displacement, $x$ metres, from the origin $O$ at time $t$ seconds is given by
$$x = 2 + t - \frac { 1 } { 10 } \mathrm { e } ^ { t }$$
\begin{enumerate}[label=(\alph*)]
\item Find the distance of $P$ from $O$ when $t = 0$.
\item Find, correct to 1 decimal place, the value of $t$ when the velocity of $P$ is zero.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [6]}}