An object is at rest at #(9 ,7 ,5 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(8 ,2 ,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 · 2017-06-04T06:59:21

The time is #=2.88s#

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((8-9)^2+(2-7)^2+(6-5)^2)#

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

#=sqrt(1+25+1)#

#=sqrt27#

#=5.196m#

We apply the equation of motion

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

#u=0#

so,

#d=1/2at^2#

#a=5/4ms^-2#

#t^2=(2d)/a=(2*5.196)/(5/4)#

#=8.31s^2#

#t=sqrt(8.31)=2.88s#