An object is at rest at $(9 ,7 ,5 )$ and constantly accelerates at a rate of $5/4 m/s^2$ as it moves to point B. If point B is at $(1 ,9 ,2 )$ , 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 · 2018-03-29T07:12:54

The time is $=3.75s$

The distance $AB$ is

$AB=sqrt((1-9)^2+(9-7)^2+(2-5)^2)=sqrt(64+4+9)=sqrt(77)m$

Apply the equation of motion

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

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

The acceleration is $a=5/4ms^-2$

Therefore,

$sqrt77=0+1/2*5/4*t^2$

$t^2=8/5*sqrt77$

$t=sqrt(8/5*sqrt77)=3.75s$

An object is at rest at (9 ,7 ,5 )(9 ,7 ,5 ) and constantly accelerates at a rate of 5/4 m/s^25/4 m/s^2 as it moves to point B. If point B is at (1 ,9 ,2 )(1 ,9 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.