An object is at rest at $(5 ,2 ,8 )$ and constantly accelerates at a rate of $2/5$ $ms^-2$ as it moves to point B. If point B is at $(6 ,3 ,2 )$ , 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 · 2018-02-26T13:55:48

First we find the distance between the two points:

$d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$
$d=sqrt((6-5)^2+(3-2)^2+(2-8)^2)$
$d=sqrt(1+1+36)=sqrt(38)$

Then we know $d=ut+1/2at^2$ . We know $u=0$ , and rearranging we get:

$t=sqrt((2d)/a)=sqrt((2xxsqrt(38))/(2/5))~=5.55$ $s$

Answer 2 · 2018-02-26T13:59:52

The time is $=5.55s$

The distance $AB$ is

$AB=sqrt((6-5)^2+(3-2)^2+(2-8)^2)=sqrt(1+1+36)=sqrt(38)m$

Apply the equation of motion

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

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

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

Therefore,

$sqrt38=0+1/2*2/5*t^2$

$t^2=5*sqrt38$

$t=sqrt(5*sqrt38)=5.55s$

An object is at rest at (5 ,2 ,8 )(5 ,2 ,8 ) and constantly accelerates at a rate of 2/52/5 ms^-2ms^-2 as it moves to point B. If point B is at (6 ,3 ,2 )(6 ,3 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.