Standard +0.3 This is a straightforward second-order linear homogeneous differential equation with constant coefficients in vector form. Students solve the characteristic equation (λ² - 4λ = 0), apply initial conditions to find the velocity vector, then evaluate at the given time. The calculation involves basic exponentials and vector magnitude. While it requires multiple steps, each is routine for Further Maths students, making it slightly easier than average.
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 Q3 [7]}}