An object's two dimensional velocity is given by #v(t) = (t^2 -t +1 , t^3- 3t )#. What is the object's rate and direction of acceleration at #t=3 #?

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Answer 1 · 2017-07-23T06:12:04

The rate of acceleration is #=24.52ms^-2# in the dorection #=78.2^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(t^2-t+1, t^3-3t)#

#a(t)=v'(t)=(2t-1,3t^2-3)#

At #t=3#, the acceleration is

#a(3)=(5,24)#

The rate of acceleration is

#=||a(3)||=sqrt(5^2+24^2)=sqrt601=24.52ms^-2#

The direction is

#theta=arctan(24/5)=78.2^@#