If the position vector of the particle is given by vecr=3hatt^(2)hati+5thatj+4hatk, Find the (a) The velocity of the particle at t=3s (b) Speed of the particle at t= 3s (c) Acceleration of the particle at time t=3s

If the position vector of the particle is given by vecr=3hatt^(2)hati+

1.	If the position vector of the particle is given by vecr=3hatt^(2)hati+5thatj+4hatk, Find the (a) The velocity of the particle at t=3s (b) Speed of the particle at t= 3s (c) Acceleration of the particle at time t=3s
Answer» Solution :The velocity `VECV=(dvecr)/(dt)=(dx)/(dt)hatj+(DZ)/(dt)hatk` We obtain `vecv(t)=6thati+5hatj` The velocity has only two components `v_(x)=6t`, depending on time and `v_(y)=5` which is independent of time. The velocity at t=3s is `vecv(3)=18hati+5hatj` (b) The speed at t=3s is `v=sqrt(18^(2)+5^(2))=sqrt(349)=18.68ms^(-1)` (c) The ACCELERATION `veca` is, `veca=(d^(2)vecr)/(dt^(2))=6hati` The acceleration has only the x-component. NOTE that acceleration here is independent of t, which means `veca` is constant. EVEN at 1=3s it has same value `veca+6hati`. The velocity is non-uniform, but the acceleration is uniform (constant) in this case.