An object is at rest at #(4 ,8 ,2 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(8 ,1 ,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 · 2017-04-19T13:52:15

The time is #=3.77s#

The distance between the points #A=(x_A,y_A,z_A)# and the point #B=(x_B,y_B,z_B)# is

#AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)#

#d=AB= sqrt((8-4)^2+(1-8)^2+(7-2)^2)#

#=sqrt(4^2+(-7)^2+5^2)#

#=sqrt(16+49+25)#

#=sqrt90#

We apply the equation of motion

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

#u=0#

so,

#d=1/2at^2#

#t^2=(2d)/a=(2sqrt90)/(4/3)#

#=14.23#

#t=sqrt(14.23)=3.77s#