An object is at rest at (5 ,2 ,8 )(5,2,8) and constantly accelerates at a rate of 1/5 m/s15ms as it moves to point B. If point B is at (6 ,3 ,2 )(6,3,2), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Nov 26, 2017

The time is =7.85s=7.85s

Explanation:

The distance ABAB is

AB=sqrt((6-5)^2+(3-2)^2+(2-8)^2)=sqrt(1+1+36)=(sqrt38)mAB=(65)2+(32)2+(28)2=1+1+36=(38)m

Apply the equation of motion

s=ut+1/2at^2s=ut+12at2

The initial velocity is u=0ms^-1u=0ms1

The acceleration is a=1/5ms^-2a=15ms2

Therefore,

sqrt38=0*1/2*1/5*t^238=01215t2

t^2=10sqrt38t2=1038

t=sqrt(10sqrt38)=7.85st=1038=7.85s