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

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Answer 1 · 2017-04-29T16:05:54

The rate of acceleration is #=5.04ms^-2# in the direction #=172.6#º

Acceleration is the derivative of the velocity

#a(t)=v'(t)#

#v(t)=(t^2-3t+2,sint)#

#a(t)=v'(t)=(2t-3,cost)#

When #t=4#,

#a(4)=(2*4-3,cos4)=(5,-0.65)#

The rate of acceleration is

#||a(4)||=sqrt((5)^2+(-0.65)^2)#

#=sqrt25.42=5.04#

The direction is in the 2nd quadrant

#theta=180-arctan(0.65/5)=172.6º#