VC1.9

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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 3tj+\sin 3tk\,

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 3tj+3\cos 3tk\,

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

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}}\,

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