# An object is at rest at (1 ,6 ,9 ) and constantly accelerates at a rate of 1  ms^-2 as it moves to point B. If point B is at (6 ,4 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jun 7, 2017

The distance is $8.06$ $m$, and the object will take $4.01$ $s$ to travel this distance.

#### Explanation:

First we need to find the distance between the two points:

$s = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$

$s = \sqrt{{\left(6 - 1\right)}^{2} + {\left(4 - 6\right)}^{2} + {\left(3 - 9\right)}^{2}}$

$s = \sqrt{{\left(5\right)}^{2} + {\left(- 2\right)}^{2} + {\left(- 6\right)}^{2}}$

$s = \sqrt{25 + 4 + 36} = \sqrt{65} = 8.06$ $m$ (assuming the units are in metres)

Then we can use $s = u t + \frac{1}{2} a {t}^{2}$

The initial velocity, $u$, is equal to $0$, so the first term disappears.

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

Rearranging:

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \left(8.06\right)}{1}} = 4.01$ $s$