An object is at rest at #(1 ,3 ,5 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,6 ,4 )#, 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 · 2017-07-02T06:24:24

The time is #=2.37s#

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((3-1)^2+(6-3)^2+(4-5)^2)#

#=sqrt(2^2+3^2+1^2)#

#=sqrt(4+9+1)#

#=sqrt14#

#=3.74m#

We apply the equation of motion

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

#u=0#

so,

#d=1/2at^2#

#a=4/3ms^-2#

#t^2=(2d)/a=(2*3.74)/(4/3)#

#=5.61s^2#

#t=sqrt(5.61)=2.37s#