# An object is at rest at (8 ,3 ,3 ) and constantly accelerates at a rate of 7/4 ms^-2 as it moves to point B. If point B is at (6 ,3 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jul 8, 2017

First step is to find the distance between the two points: $r \cong 2.24$ $m$

Second step is to find the time taken: $t = 1.6$ $s$

#### Explanation:

To find the distance between two points in 3D space we have:

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

$= \sqrt{{\left(6 - 8\right)}^{2} + {\left(3 - 3\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{{\left(- 2\right)}^{2} + {\left(0\right)}^{2} + {\left(1\right)}^{2}}$

Therefore:

$r = \sqrt{5} \cong 2.24$

Now to find the time taken, we can use:

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

Since the object is initially at rest, the $u t$ term goes to $0$.

Rearranging:

$t = \sqrt{\frac{2 d}{a}}$

Note that I have used both $d$ and $r$ because of the way I remember formulae, but we're talking about the same distance - the distance between the points.

$t = \sqrt{\frac{2 \times 2.24}{\frac{7}{4}}} = 1.6$ $s$