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

May 7, 2016

The distance between the points is $7.48$ $m$.

The time taken is $t = \sqrt{\frac{d}{a}} = \sqrt{\frac{7.48}{0.8}} \approx 3.06$ $s$.

#### Explanation:

First find the distance between the points:

$l = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$= \sqrt{{\left(6 - 2\right)}^{2} + {\left(6 - 4\right)}^{2} + {\left(7 - 1\right)}^{2}} = \sqrt{16 + 4 + 36}$
$= \sqrt{56} \approx 7.48$ $m$

An acceleration of $\frac{4}{5}$ can be expressed as $0.8$.

$d = u t + a {t}^{2}$ can be rearranged to $t = \sqrt{\frac{d}{a}}$ when $u = 0$ (which it is in this case).