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=5 #?

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Answer 1 · 2017-03-03T09:23:20

#a(5) = -70 hat i + 28 hat j #

#v (t) = 5t-t^3 hati + 3t^2-2t hatj#

#a(t) = (dv)/dt#

#a(t) = 5-3t^2 hat i + 6t-2 hat j #

#a(5) = 5-3xx5^2 hat i + 6xx5-2 hat j #

#a(5) = 5-75 hat i + 30-2 hat j #

#a(5) = -70 hat i + 28 hat j #

Here, #hat i # denotes #x# direction and #hatj# denotes #y# direction.