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