# What is the speed of an object that travels from ( 8 , 4, 1 )  to ( 6 , 0, ,2 )  over 2 s?

Mar 18, 2016

$v = \sqrt{6} \frac{\text{ ""unit}}{s}$

#### Explanation:

${P}_{1} \left(8 , 4 , 1\right) \text{ } {P}_{2} \left(6 , 0 , 2\right)$
${P}_{\text{1x"=8" "P_"2x"=6" }} \Delta {P}_{x} = 6 - 8 = - 2$
${P}_{\text{1y"=4" "P_"2y"=0" }} \Delta {P}_{y} = 0 - 4 = - 4$
${P}_{\text{1z"=1" "P_"2z"=2" }} \Delta {P}_{z} = 2 - 1 = 2$
$\text{ distance between the point of " P_1" and " P_2" is:}$
$\Delta x = \sqrt{{\left(\Delta {P}_{x}\right)}^{2} + {\left(\Delta {P}_{y}\right)}^{2} + {\left(\Delta {P}_{z}\right)}^{2}}$

$\Delta x = \sqrt{{\left(- 2\right)}^{2} + {\left(- 4\right)}^{2} + {2}^{2}} = \sqrt{4 + 16 + 4} = \sqrt{24}$

$v = \frac{\Delta x}{t}$

$v = \frac{\sqrt{24}}{2}$
$v = \frac{\sqrt{4 \cdot 6}}{2}$
$v = \frac{\cancel{2} \cdot \sqrt{6}}{\cancel{2}}$

$v = \sqrt{6} \frac{\text{ ""unit}}{s}$