Standard +0.3 This is a straightforward second-order vector differential equation requiring standard solution of the auxiliary equation (giving r = 0, 4), application of initial conditions, and evaluation at a specific time. The homogeneous nature and simple exponential solution make it easier than average, though the vector context and logarithmic time value add minor complexity.
3. At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf { r }\) metres, relative to a fixed origin. The particle moves in such a way that
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$
At \(t = 0 , P\) is moving with velocity ( \(8 \mathbf { i } - 6 \mathbf { j }\) ) \(\mathrm { m } \mathrm { s } ^ { - 1 }\).
Find the speed of \(P\) when \(t = \frac { 1 } { 2 } \ln 2\).
3. At time $t$ seconds, the position vector of a particle $P$ is $\mathbf { r }$ metres, relative to a fixed origin. The particle moves in such a way that
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$
At $t = 0 , P$ is moving with velocity ( $8 \mathbf { i } - 6 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$.\\
Find the speed of $P$ when $t = \frac { 1 } { 2 } \ln 2$.\\
\hfill \mbox{\textit{Edexcel M5 2002 Q3 [7]}}