The position of a particle is given by
`r = 3t hati +2t^(2) hatj +8 hatk`
where, t is in seconds and the coefficients have the proper units for r to be in meters.
(i) Find v (t) and a(t) of the particles.
(ii) Find the magnitude and direction of v(t) and a(t) at `t = 1s`.

Explanation

Position of particle, `r = 3t hati +2t^(2) hatj +8 hatk`
The formula used
(i) `v(t) = (dr)/(dt)` and `a(t) =(dv)/(dt)`
(ii) `v(t) = sqrt(v_(x)^(2)+v_(y)^(2))` and `theta = tan^(-1) ((v_(y))/(v_(x)))`
`:. v(t) = (d)/(dt) (3 thati +2t^(2) hatj +8hatk) = 3hati +4thatj`
`a(t) = (dv)/(dt) = 4 hatj`
At `t = 1s`
Velocity, `v(t) = 3hati +4t hatj`
`:. v = sqrt((3)^(2)+(4)^(2)) = 5 ms^(-1)`
`rArr` Direction of `v(t) = theta = tan^(-1) ((4)/(3)) = 53^(@)` with X-axis
Direction of a(t), at `t = 1s`
`theta' = tan^(-1) ((a_(y))/(a_(x))) = tan^(-1) ((4)/(0)) = tan^(-1) (oo) = 90^(@)` with X-axis