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

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Answer 1 · 2017-03-29T08:30:13

2nd quandrant (negative x and positive y)

When you put 3 (for t), you will get v(t) = (-5, 9). The distance (if speed is expressed as m/s and time is expressed as s) 30.89 meters.

Note that the units of time and velocity were not provided in the original question.

Answer 2 · 2017-03-29T08:30:23

The rate of acceleration is #=6.18ms^-2# in the direction of #=104#º

#v(t)=(sqrt(t-2)-2t,t^2)#

The acceleration is the derivative of the velocity

#a(t)=v'(t)=(dv)/dt#

#=(1/2*1/sqrt(t-2)-2,2t)#

Therefore,

#a(3)=(1/2-2,6)=(-3/2,6)#

The rate of acceleration is

#=sqrt((-3/2)^2+6^2)#

#=sqrt(9/4+36)#

#=sqrt153/2#

#=6.18ms^-2#

#tan theta=6/(-3/2)=-4#

The angle is in the 2nd quadrant and is

#theta=arctan(-4)=104#º