An object is at rest at #(3 ,8 ,8 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(1 ,9 ,2 )#, 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 · 2017-08-11T12:02:49

The time is #=2.86s#

The distance #AB# is

#AB=sqrt((1-3)^2+(9-8)^2+(2-8)^2)=sqrt((-2)^2+(1)^2+(-6)^2)#

#=sqrt(4+1+36)=sqrt41m#

The acceleration is #a=5/4ms^-2#

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

We apply the equation of motion

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

So,

#sqrt41=0+1/2*(5/4)^2*t^2#

#t^2=32/25sqrt41#

#t=sqrt(32/25sqrt41)=2.86s#