An object is at rest at #(2 ,1 ,3 )# and constantly accelerates at a rate of #7/5 m/s^2# as it moves to point B. If point B is at #(9 ,3 ,5 )#, 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 · 2018-03-14T07:29:41

The time is #=3.28s#

The distance #AB# is

#AB=sqrt((9-2)^2+(3-1)^2+(5-3)^2)=sqrt(49+4+4)=sqrt(57)m#

Apply the equation of motion

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

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

The acceleration is #a=7/5ms^-2#

Therefore,

#sqrt57=0+1/2*7/5*t^2#

#t^2=10/7*sqrt57#

#t=sqrt(10/7*sqrt57)=3.28s#