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 #?

1 Answer
Jan 30, 2017

Answer:

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.