Easy -1.8 This is a straightforward differentiation exercise requiring only the application of the chain rule to find velocity from position, followed by direct substitution of t=0. It's a 1-mark multiple-choice question testing basic recall of differentiation and vector notation, with no problem-solving or conceptual challenge beyond routine mechanics.
A particle moves in a circular path so that at time \(t\) seconds its position vector, \(\mathbf{r}\) metres, is given by
$$\mathbf{r} = 4\sin(2t)\mathbf{i} + 4\cos(2t)\mathbf{j}$$
Find the velocity of the particle, in m s\(^{-1}\), when \(t = 0\)
Circle your answer.
[1 mark]
\(8\mathbf{i}\) \quad \(-8\mathbf{j}\) \quad \(8\mathbf{j}\) \quad \(8\mathbf{i} - 8\mathbf{j}\)
A particle moves in a circular path so that at time $t$ seconds its position vector, $\mathbf{r}$ metres, is given by
$$\mathbf{r} = 4\sin(2t)\mathbf{i} + 4\cos(2t)\mathbf{j}$$
Find the velocity of the particle, in m s$^{-1}$, when $t = 0$
Circle your answer.
[1 mark]
$8\mathbf{i}$ \quad $-8\mathbf{j}$ \quad $8\mathbf{j}$ \quad $8\mathbf{i} - 8\mathbf{j}$
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q1 [1]}}