An object's two dimensional velocity is given by #v(t) = ( t^2, t-t^2sin(pi/3)t)#. What is the object's rate and direction of acceleration at #t=4 #?

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Answer 1 · 2016-12-26T19:33:37

Please see the explanation.

#v_x(t) = t^2#

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

#v_y(t) = t - t^2sin((pit)/3)#

#(dv_y(t))/dt = 1 - 2tsin((pit)/3) - (pit^2)/3cos((pit)/3)#

The magnitude of the acceleration is

#a = sqrt((2t)^2 + (1 - 2tsin((pit)/3) - (pit^2)/3cos((pit)/3))^2) #

I used WolframAlpha to evaluate the above at t = 4:

#a ~~ 9.43#

The direction is along the unit vector:

#(1/9.43)(64hati - 4.99hatj)#