An object is at rest at $(9 ,7 ,3 )$ and constantly accelerates at a rate of $1/4 m/s^2$ as it moves to point B. If point B is at $(1 ,3 ,2 )$ , 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 · 2018-01-08T13:37:29

The time is $=8.49s$

The distance $AB$ is

$AB=sqrt((1-9)^2+(3-7)^2+(2-3)^2)=sqrt(64+16+1)=sqrt81=9m$

Apply the equation of motion

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

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

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

Therefore,

$9=0+1/2*1/4*t^2$

$t^2=9*8$

$t=sqrt(9*8)=8.49s$

An object is at rest at (9 ,7 ,3 )(9 ,7 ,3 ) 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 (1 ,3 ,2 )(1 ,3 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.