The position of a particle is given by `r=3.0thati + 2.0 t^(2)hatj + 5.0 hatk` . Where t is in seconds and the coefficients have the proper units for r to be in metres, (a) Find v(t) and aft) of the particle, (b) Find the magnitude and direction ofv(t) at t = 1.0 s.

Explanation

`v(t) (dr)/(dt) =d/t(3.0 hati +2.0t^(2)hatj + 5.0hatk)=3.0hati + 4.0hatj`
`a(t) =(dv)/(dt) =+4.0 hatj, a=4.0 ms^(-2)` along y-direction
(b) At t=1.0 s, `v=3.0hati + 4.0hatj`
It.s magnitude is `v=sqrt(3^(2) + 4^(2)) =5.0 ms^(-1)`
and direction is
`theta = tan^(-1)(v_(y)/v_(x)) = tan^(-1)(4/3) = 53^(@)` with x-axis.