VC1.9

Let r be the position vector of the position $P(x,y,z)\,$ of the particle at time t.Then,we have

$r=xi+yj+zk=e^{-t}i+2\cos 3t j+\sin 3t k\,$

If v and a be the velocity and acceleration respectively of the particle at time t,

Then

$v=\frac{dr}{dt}=-e^{-t}i-6\sin 3t j+3\cos 3t k\,$

$a=\frac{d^2 r}{dt^2}=e^{-t}i-18\cos 3t j-9\sin 3t k\,$

Hence, at t=0 $v=-i+3k,a=-i-18j\,$

Now the magnitude of velocity = $\sqrt{1+9}=\sqrt{10}\,$

The magnitude of acceleration = $\sqrt{1+18^{2}}=\sqrt{325}\,$