The position of a particle is given by `r=3.0that(i)+2.0t^(2)hat(j)+5.0hat(k)`. Where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0s.

Explanation

`v(t)(dr)/(dt)=(d)/(t)(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 = 4.0 m s^(-2)` along y direction
(b) `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^(0)` with x-axis.