An object is at rest at (5 ,2 ,1 )(5,2,1) and constantly accelerates at a rate of 2/3 m/s23ms as it moves to point B. If point B is at (4 ,9 ,8 )(4,9,8), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Apr 2, 2018

The time is =5.46s=5.46s

Explanation:

The distance ABAB is

AB=sqrt((4-5)^2+(9-2)^2+(8-1)^2)=sqrt(1+49+49)=sqrt(99)mAB=(45)2+(92)2+(81)2=1+49+49=99m

Apply the equation of motion

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

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

The acceleration is a=2/3ms^-2a=23ms2

Therefore,

sqrt99=0+1/2*2/3*t^299=0+1223t2

t^2=3*sqrt99t2=399

t=sqrt(3*sqrt99)=5.46st=399=5.46s