An object is at rest at (8 ,4 ,2 ) and constantly accelerates at a rate of 4/3 m/s as it moves to point B. If point B is at (3 ,1 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 6, 2017

sqrt(21/2)s

Explanation:

let initial point A be (8,4,2) and final point B be (3,1,6)

by distance formula,

= sqrt[(8-3)^2 + (4-1)^2 + (2-6)^2]

therefore distance 's' between A and B is sqrt50

So, s = sqrt50 = 5sqrt2 = 5*1.41 = 7m approx

also,
as object is at rest and starts constantly accelerating,
initial velocity ' u' = 0

By using equation,
s = ut + 1/2at^2

as u = 0, s = 1/2at^2

7 = 1/2*4/3t^2

t = sqrt(21/2)s