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

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Answer 1 · 2016-09-26T22:36:06

The vector is #a(t) = ((1/2)(t - 2)^(-1/2) - 1, 2t)# Substitute 3 for t, and then compute the magnitude and direction as you would any vector.

#a(3) = ((1/2)(3 - 2)^(-1/2) - 1, 2(3))#

#a(3) = (-1/2, 6)#

The rate of acceleration is the magnitude:

#|a(3)| = sqrt((-1/2)^2 + 6^2)#

#a ~~ 6.02 (units)/s^2#

#theta = tan^-1(6/(-1/2))#

#theta ~~ -85.2°#