An object is at rest at #(3 ,8 ,8 )# and constantly accelerates at a rate of #7/5 m/s^2# as it moves to point B. If point B is at #(2 ,9 ,1 )#, 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 · 2017-06-18T17:13:54

The time is #=3.2s#

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((2-3)^2+(9-8)^2+(1-8)^2)#

#=sqrt(1^2+1^2+7^2)#

#=sqrt(1+1+49)#

#=sqrt51#

#=7.14m#

We apply the equation of motion

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

#u=0#

so,

#d=1/2at^2#

#a=7/5ms^-2#

#t^2=(2d)/a=(2*7.14)/(7/5)#

#=10.2s^2#

#t=sqrt(10.2)=3.2s#