An object is at rest at $(2 ,1 ,3 )$ and constantly accelerates at a rate of $1 m/s^2$ as it moves to point B. If point B is at $(3 ,7 ,5 )$ , 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 · 2018-04-04T06:11:58

The time is $=3.58s$

The distance $AB$ is

$AB=sqrt((3-2)^2+(7-1)^2+(5-3)^2)=sqrt(1+36+4)=sqrt(41)m$

Apply the equation of motion

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

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

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

Therefore,

$sqrt41=0+1/2*1*t^2$

$t^2=2*sqrt41$

$t=sqrt(2*sqrt41)=3.58s$

An object is at rest at (2 ,1 ,3 )(2 ,1 ,3 ) and constantly accelerates at a rate of 1 m/s^21 m/s^2 as it moves to point B. If point B is at (3 ,7 ,5 )(3 ,7 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.