An object is at rest at #(2 ,7 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,1 ,6 )#, 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 · 2017-05-26T08:25:36

The time is #=3.44s#

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((3-2)^2+(1-7)^2+(6-1)^2)#

#=sqrt(1^2+6^2+5^2)#

#=sqrt(1+36+25)#

#=sqrt62#

#=7.874m#

We apply the equation of motion

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

#u=0#

so,

#d=1/2at^2#

#t^2=(2d)/a=(2*7.874)/(4/3)#

#=11.81#

#t=sqrt(11.81)=3.44s#