An object is at rest at $(5 ,2 ,1 )$ and constantly accelerates at a rate of $2/3 m/s$ as it moves to point B. If point B is at $(4 ,9 ,8 )$ , 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 · 2018-04-02T06:50:49

The time is $=5.46s$

The distance $AB$ is

$AB=sqrt((4-5)^2+(9-2)^2+(8-1)^2)=sqrt(1+49+49)=sqrt(99)m$

Apply the equation of motion

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

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

The acceleration is $a=2/3ms^-2$

Therefore,

$sqrt99=0+1/2*2/3*t^2$

$t^2=3*sqrt99$

$t=sqrt(3*sqrt99)=5.46s$

An object is at rest at (5 ,2 ,1 )(5 ,2 ,1 ) and constantly accelerates at a rate of 2/3 m/s2/3 m/s as it moves to point B. If point B is at (4 ,9 ,8 )(4 ,9 ,8 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.