# An object is at rest at (1 ,2 ,1 ) and constantly accelerates at a rate of 1/3 ms^-2 as it moves to point B. If point B is at (4 ,4 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jan 11, 2018

The distance between the two points is $5.4$ $m$ and the time taken to move from one to the other is $5.7$ $s$.

#### Explanation:

First step is to find the distance between the two points:

$s = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$= \sqrt{{\left(4 - 1\right)}^{2} + {\left(4 - 2\right)}^{2} + {\left(5 - 1\right)}^{2}} = \sqrt{{3}^{2} + {2}^{2} + {4}^{2}}$
$= \sqrt{9 + 4 + 16} = \sqrt{29} = 5.4$ $m$

Now we know:

$u = 0$ $m {s}^{-} 1$ (since the object was initially at rest)
$a = \frac{1}{3}$ $m {s}^{-} 2$
$s = 5.4$ $m$ (some people use $d$ for distance)
t=? $s$

We have the equation of motion:

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

The $u t$ term goes to 0 because $u = 0$, which makes life simpler:

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

Rearranging to make $t$ the subject:

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \times 5.4}{\frac{1}{3}}} = \sqrt{32.4} = 5.7$ $s$