Question

An object's velocity is given by `v(t) = (t^2 -t +1 , t^3- 3t )`. What is the object's rate and direction of acceleration at `t=2`?

Answer 1

`v_x(t) = t^2-t+1`

`a_x(t) = dotv_x(t) = 2t-1`

`:. a_x(2) = 3`

`v_y(t) = t^3-3t`
`a_y(t) = dotv_y(t) =3t^2-3`
`:.a_y(2) = 9`

Hence, `|a| = sqrt(3^2+9^2)=sqrt90 = 3sqrt10`

And direction is given as:
`tantheta = 9/2`