An object's two dimensional velocity is given by #v(t) = ( 5t-t^3 , 3t^2-2t)#. What is the object's rate and direction of acceleration at #t=7 #?

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Answer 1 · 2017-05-14T07:15:16

The rate of acceleration is #=147.53ms^-1# in the direction of #=164.27º# anticlockwise from the x-axis.

The acceleration is the derivative of the velocity.

#v(t)=(5t-t^3, 3t^2-2t)#

#a(t)=v'(t)=(5-3t^2, 6t-2)#

Therefore,

#a(7)=(5-3*49, 6*7-2)=(-142,40)#

The rate of acceleration is

#||a(4)||=sqrt(142^2+40^2)#

#=sqrt21764#

#=147.53ms^-2#

The direction is

#theta=arctan(-40/142)#

#theta# lies in the 2nd quadrant

#theta=180-15.73=164.27#