Question

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

Answer 1

We take the first derivative with respect to time to get the component (rectangular) form of the acceleration, then change this to standard (polar) form, to get the acceleration as `5m/s^2`
in a direction of 36.9° south of east.

Explanation:

Since the acceleration is the first derivative with respect to time, the acceleration must be

#a(t) = (6t-2, -3)

and at time `t=1` this becomes

`a=(4,-3)`

To change to standard form, we use the following operations:

`|a|=sqrt(a_x^2+a_y^2) = sqrt(4^2+(-3)^3) = 5 m/s^2`

The direction is given by `tan^-1 (a_y/a_x) = tan^-1(-3/4)=-36.9°`

So, the acceleration is `5m/s^2` in a direction of 36.9° south of east.