An object is at rest at $(2 ,8 ,6 )$ and constantly accelerates at a rate of $1/4 m/s^2$ as it moves to point B. If point B is at $(3 ,2 ,7 )$ , 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 · 2017-11-24T14:02:59

The time is $=7.02s$

The distance $AB$ is

$AB=sqrt((3-2)^2+(2-8)^2+(7-6)^2)=sqrt(1+36+1)=sqrt38 m$

Apply the equation of motion

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

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

The final velocity is $v=?$

The acceleration is $a=1/4ms^-2$

Therefore,

$sqrt38=0*1/2*1/4*t^2$

$t^2=8sqrt38$

$t=sqrt(8sqrt38)=7.02s$

An object is at rest at (2 ,8 ,6 )(2 ,8 ,6 ) and constantly accelerates at a rate of 1/4 m/s^21/4 m/s^2 as it moves to point B. If point B is at (3 ,2 ,7 )(3 ,2 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.