An object is at rest at $(4 ,7 ,3 )$ and constantly accelerates at a rate of $7/5 m/s^2$ as it moves to point B. If point B is at $(5 ,6 ,9 )$ , 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 · 2018-01-03T07:25:52

The time is $=3.0s$

The distance $AB$ is

$AB=sqrt((5-4)^2+(6-7)^2+(9-3)^2)=sqrt(1+1+36)=(sqrt38)m$

Apply the equation of motion

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

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

The acceleration is $a=7/5ms^-2$

Therefore,

$sqrt(38)=0+1/2*7/5*t^2$

$t^2=10/7sqrt38$

$t=sqrt(10/7sqrt38)=3.0s$

An object is at rest at (4 ,7 ,3 )(4 ,7 ,3 ) and constantly accelerates at a rate of 7/5 m/s^27/5 m/s^2 as it moves to point B. If point B is at (5 ,6 ,9 )(5 ,6 ,9 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.