Given that the acceleration vector is a(t) = (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k , the initial velocity is v(0) = i + k , and the initial position vector is r(0) = i+j+k , compute:

A. The velocity vector

B. The position vector

Solution Preview

a(t) = (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k

initial velocity is v(0) = i + k , and the initial position vector is r(0) = i+j+k

We know that acceleration = dV/dt, the Rate of change of velocity wrt Time

So V(t) = integral [a*dt] = integral[ (-9cos(-3t)) i + (-9sin(-3t)) j + (-2t) k]*dt

= -3Sin(3t) i - 3Cos(3t) j - t^2 k + C --------A

When t= 0, We get V(0) = -3Sin(0) ...

Solution Summary

The problem is solved giving all necessary mathematical steps. It will enable you to do similar problems yourself.

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