An object is at rest at #(2 ,8 ,6 )# and constantly accelerates at a rate of #5/3 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 · 2017-08-15T17:36:51

The time is #=2.11s#

The distance #AB# is

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

#=sqrt(1+36+1)=sqrt38m#

The acceleration is #a=5/3ms^-2#

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

We apply the equation of motion

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

So,

#sqrt38=0+1/2*(5/3)^2*t^2#

#t^2=18/25sqrt38#

#t=sqrt(18/25sqrt38)=2.11s#