| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Moderate -0.3 This is a straightforward second-order vector differential equation with constant coefficients. The auxiliary equation method is routine (giving r=0,2), and applying the two initial conditions to find constants requires only basic substitution. While it involves vectors, the technique is standard M5 material with no conceptual challenges. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration4.10b Model with differential equations: kinematics and other contexts |
A particle moves in a plane in such a way that its position vector $\mathbf{r}$ metres at time $t$ seconds satisfies the differential equation
$$\frac{d^2\mathbf{r}}{dt^2} = \frac{d\mathbf{r}}{dt} - 2\mathbf{0}$$
When $t = 0$, the particle is at the origin and is moving with velocity $(4\mathbf{i} + 2\mathbf{j})$ m s$^{-1}$.
Find $\mathbf{r}$ in terms of $t$.
(7 marks)
---\begin{enumerate}
\item A particle moves in a plane in such a way that its position vector $\mathbf { r }$ metres at time $t$ seconds satisfies the differential equation
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$
When $t = 0$, the particle is at the origin and is moving with velocity $( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
Find $\mathbf { r }$ in terms of $t$.\\
\hfill \mbox{\textit{Edexcel M5 2013 Q1 [7]}}