An object's velocity is given by #v(t) = (2t^2 +t +1 , sin2t )#. What is the object's rate and direction of acceleration at #t=1 #?

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Answer 1 · 2018-06-02T08:14:21

The rate of acceleration is #=5.069ms^-2# in the direction #=9.43^@# clockwise from the x-axis.

The acceleration is the derivative of the velocity.

#v(t)=(2t^2+t+1, sin2t)#

#a(t)=v'(t)=(4t+1, 2cos2t)#

Therefore,

When #t=1#

#a(1)=(5,2cos2)= (5, -0.83)#

The rate of acceleration is

#||a(1)||=sqrt(5^2+(-0.83)^2)=5.069ms^-2#

The direction is

#theta=arctan(-0.83/5)=9.43^@# clockwise from the x-axis