# An object is at rest at (2 ,1 ,6 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (3 ,2 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

May 21, 2017

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \times 1.73}{0.25}} = 3.72$ $s$

#### Explanation:

In a 3-dimensional coordinate system, the distance between two points is given by:

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

$r = \sqrt{{\left(3 - 2\right)}^{2} + {\left(2 - 1\right)}^{2} + {\left(7 - 6\right)}^{2}} = \sqrt{3} = 1.73$ $m$

We can use the equation of motion:

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

Since the initial velocity $u = 0$, we can ignore the first term, and rearrange $s = \frac{1}{2} a {t}^{2}$ to make $t$ the subject.

Note that the term 'r' I was using in the distance calculation above is the same as the distance 's' that the object moves. I don't want the symbol change to be confusing.

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \times 1.73}{0.25}}$

(I used 0.25 rather than 1/4 for ease of writing, but of course it means the same thing)

Therefore $t = 3.72$ $s$.