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=5 #?

1 Answer
Jun 24, 2017

Answer:

The acceleration is #28.2# #ms^-2# at #-84^o#: that is, at an angle in the third quadrant, downward and to the right.

Explanation:

The acceleration is the rate of change of the velocity, and is found as the first derivative of the velocity. We can treat the #x# and #y# directions independently.

#v_x=3t^2-2t#

#(dv_x)/dt=6t-2#

#v_y=1-3t#

#(dv_y)/dt=-3#

We can recombine these dimensions:

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

Interestingly, the acceleration in the #y# direction is constant, while the acceleration in the #x# direction changes with time.

At #t=5# #s#, we substitute in 5, and get:

#a(t)=(6(5)-2,-3)=(28,-3)#

We can use Pythagoras' theorem to find the magnitude and the tangent to find the direction of the resultant acceleration vector.

#r=sqrt(x^2+y^2)=sqrt(28^2+(-3)^2)=sqrt(793)=28.2# #ms^-2#

#tan \theta = 28/(-3) -> \theta = -84^o#