An object's velocity is given by #v(t) = (2t^2 -t +2 , sint )#. What is the object's rate and direction of acceleration at #t=4 #?

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Answer 1 · 2017-05-06T09:46:33

The rate of acceleration is #=15.01ms^-2# in the direction of #=177.5º#

The acceleration is the derivative of the velocity.

#v(t)=(2t^2-t+2,sint)#

#a(t)=v'(t)=(4t-1,cost)#

Therefore,

#a(4)=(15,cos4)=(15,-0.65)#

The rate of acceleration is

#||a(4)||=sqrt(15^2+(0.65)^2)#

#=sqrt225.4#

#=15.01ms^-2#

The direction is

#theta=arctan(-0.65/15)#

#theta# lies in the 2nd quadrant

#theta=180-2.5=177.5º#