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

1 Answer
Mar 12, 2017

Answer:

The rate of acceleration is #=sqrt(13-3/sqrta+1/(4a))# in the direction #=arctan((4sqrta)/(6sqrta-1))#

Explanation:

#v(t)=(3t-sqrtt,2t-4)#

The acceleration is the derivative of the velocity

#v'(t)=(3-1/(2sqrtt),2)#

Therefore,

#v'(a)=(3-1/(2sqrta),2)#

The rate of acceleration is

#||v'(a)||=sqrt((3-1/(2sqrta))^2+ 4)#

#=sqrt(9-3/sqrta+1/(4a)+4)#

#=sqrt(13-3/sqrta+1/(4a))#

The direction is

#=arctan(2/(3-1/(2sqrta)))#

#=arctan((4sqrta)/(6sqrta-1))#