An object is at rest at #(2 ,8 ,6 )# and constantly accelerates at a rate of #1/4 m/s^2# as it moves to point B. If point B is at #(3 ,2 ,7 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2017-11-24T14:02:59

The time is #=7.02s#

The distance #AB# is

#AB=sqrt((3-2)^2+(2-8)^2+(7-6)^2)=sqrt(1+36+1)=sqrt38 m#

Apply the equation of motion

#s=ut+1/2at^2#

The initial velocity is #u=0ms^-1#

The final velocity is #v=?#

The acceleration is #a=1/4ms^-2#

Therefore,

#sqrt38=0*1/2*1/4*t^2#

#t^2=8sqrt38#

#t=sqrt(8sqrt38)=7.02s#