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 metres.
(i) Find v (t) and a(t) of the particles.
(ii) Find the magnitude and direction of v(t) and a(t) at `1 = 1s`.

Options

A. `(2.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
B. `(3.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
C. `(3.0hat(i) + 4.0that(j))` , `a=2 ms^(-2)` `5.0ms^(-1)` , `53^(@)` with x-axis`
D. `(3.0hat(i) + 4.0that(j))` , `a=4.0ms^(-2)` `5.0ms^(-1)` , `45^(@)` with x-axis`

Answer

B

Explanation

`v(t) = (dr)/(dt)= d/(dt)(3.0that(i) + 2.0t^(2)hat(j)+5.0hat(k))`
` =(3.0hat(i) + 4.0that(j))`
a(t) = `(dv)/(dt) = +4.0hat(j)`
`a=4.0ms^(-2)` along y-direction
At t-1.0s, `v=3.0hat(i) + 4.0hat(j)`
It's magnitude is `v=sqrt(3^(2)+4^(2)) = 5.0ms^(-1)` and direction is
`theta=tan^(-1)(v_(y)/v_(x))=tan^(-1)(4/3)=~ 53^(@)` with x-axis.