An object is at rest at #(1 ,7 ,2 )# and constantly accelerates at a rate of #1 m/s^2# as it moves to point B. If point B is at #(3 ,7 ,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 · 2016-12-22T06:34:00

The answer is #=2.4s#

The distance between 2 points A#(x_A,y_A,z_A)# and

B#(x_B,y_B, z_B)# is

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

Here we have, A#(1,7,2)# and B#(3,7,4)#

so,

#d=sqrt((3-1)^2+(7-7)^2+(4-2)^2)#

#=sqrt(4+4)=sqrt8m#

We use the equation

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

As, the object is initially at rest #u=0#

#a=1ms^(-2)#

Then,

#sqrt8=1/2*1*t^2#

#t^2=2sqrt8#

#t=sqrt(2sqrt8)=2.4s#