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=5 #?

Question

1399 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-21T11:56:30

The rate of acceleration is #=21.07ms^-2# in the direction of #=4.6^@# 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)#

When #t=5#

#a(5)=(21,2cos10)=(21,-1.68)#

The rate of acceleration is

#||a(5)||=sqrt(21^2+(-1.68)^2)=sqrt(443.8)=21.07#

The direction is

#theta=arctan(-1.68/21)=4.6^@# clockwise from the x-axis