# What is the speed of an object that travels from ( 7 , -4, 3 )  to ( -2 , 4, 9 )  over 4 s?

Apr 28, 2016

$s = \frac{d}{t} = \frac{13.45 m}{4 s} = 3.36$ $m {s}^{-} 1$

#### Explanation:

First find the distance between the points, assuming distances are in metres:

$r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$= \sqrt{{\left(\left(- 2\right) - 7\right)}^{2} + {\left(4 - \left(- 4\right)\right)}^{2} + {\left(9 - 3\right)}^{2}}$
$= \sqrt{- {9}^{2} + {8}^{2} + {6}^{2}} = \sqrt{81 + 64 + 36} = \sqrt{181} \approx 13.45$ $m$

Then speed is just distance divided by time:

$s = \frac{d}{t} = \frac{13.45}{4} = 3.36$ $m {s}^{-} 1$