# 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.

Feb 26, 2018

First we find the distance between the two points:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$d = \sqrt{{\left(6 - 5\right)}^{2} + {\left(3 - 2\right)}^{2} + {\left(2 - 8\right)}^{2}}$
$d = \sqrt{1 + 1 + 36} = \sqrt{38}$

Then we know $d = u t + \frac{1}{2} a {t}^{2}$. We know $u = 0$, and rearranging we get:

$t = \sqrt{\frac{2 d}{a}} = \sqrt{\frac{2 \times \sqrt{38}}{\frac{2}{5}}} \cong 5.55$ $s$

Feb 26, 2018

The time is $= 5.55 s$

#### Explanation:

The distance $A B$ is

$A B = \sqrt{{\left(6 - 5\right)}^{2} + {\left(3 - 2\right)}^{2} + {\left(2 - 8\right)}^{2}} = \sqrt{1 + 1 + 36} = \sqrt{38} m$

Apply the equation of motion

$s = u t + \frac{1}{2} a {t}^{2}$

The initial velocity is $u = 0 m {s}^{-} 1$

The acceleration is $a = \frac{2}{5} m {s}^{-} 2$

Therefore,

$\sqrt{38} = 0 + \frac{1}{2} \cdot \frac{2}{5} \cdot {t}^{2}$

${t}^{2} = 5 \cdot \sqrt{38}$

$t = \sqrt{5 \cdot \sqrt{38}} = 5.55 s$