An object is at rest at $(2 ,4 ,1 )$ and constantly accelerates at a rate of $4/5 m/s$ as it moves to point B. If point B is at $(3 ,6 ,7 )$ , 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 · 2018-04-12T08:20:06

The time is $=4.0s$

The distance $AB$ is

$AB=sqrt((3-2)^2+(6-4)^2+(7-1)^2)=sqrt(1+4+36)=sqrt(41)m$

Apply the equation of motion

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

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

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

Therefore,

$sqrt41=0+1/2*4/5*t^2$

$t^2=5/2*sqrt41$

$t=sqrt(5/2*sqrt41)=4.0s$

An object is at rest at (2 ,4 ,1 )(2 ,4 ,1 ) and constantly accelerates at a rate of 4/5 m/s4/5 m/s as it moves to point B. If point B is at (3 ,6 ,7 )(3 ,6 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.