An object is at rest at #(5 ,2 ,8 )# and constantly accelerates at a rate of #2/5# #ms^-2# as it moves to point B. If point B is at #(6 ,3 ,2 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

2 Answers
Feb 26, 2018

First we find the distance between the two points:

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#
#d=sqrt((6-5)^2+(3-2)^2+(2-8)^2)#
#d=sqrt(1+1+36)=sqrt(38)#

Then we know #d=ut+1/2at^2#. We know #u=0#, and rearranging we get:

#t=sqrt((2d)/a)=sqrt((2xxsqrt(38))/(2/5))~=5.55# #s#

Feb 26, 2018

Answer:

The time is #=5.55s#

Explanation:

The distance #AB# is

#AB=sqrt((6-5)^2+(3-2)^2+(2-8)^2)=sqrt(1+1+36)=sqrt(38)m#

Apply the equation of motion

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

The initial velocity is #u=0ms^-1#

The acceleration is #a=2/5ms^-2#

Therefore,

#sqrt38=0+1/2*2/5*t^2#

#t^2=5*sqrt38#

#t=sqrt(5*sqrt38)=5.55s#