Standard +0.3 This is a straightforward application of integrating factor method to a vector differential equation with standard exponential form, followed by one integration to find position. The vector notation adds minimal complexity since each component is solved independently using the same routine technique. Slightly above average difficulty due to the vector context and two-stage process, but remains a standard textbook exercise requiring no novel insight.
2. The velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\) of a particle \(P\) at time \(t\) seconds satisfies the vector differential equation
$$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$
The position vector of \(P\) at time \(t\) seconds is \(\mathbf { r }\) metres.
Given that at \(t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )\) and \(\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )\), find \(\mathbf { r }\) at time \(t\) seconds.
2. The velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ of a particle $P$ at time $t$ seconds satisfies the vector differential equation
$$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 4 \mathbf { v } = \mathbf { 0 }$$
The position vector of $P$ at time $t$ seconds is $\mathbf { r }$ metres.\\
Given that at $t = 0 , \mathbf { r } = ( \mathbf { i } - \mathbf { j } )$ and $\mathbf { v } = ( - 8 \mathbf { i } + 4 \mathbf { j } )$, find $\mathbf { r }$ at time $t$ seconds.\\
\hfill \mbox{\textit{Edexcel M5 2008 Q2 [7]}}