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

Explanation

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.