# What is the speed of an object that travels from (9,-6,1)  to (-1,5,-1 )  over 4 s?

Jan 19, 2016

$\textcolor{g r e e n}{\text{speed "=15/4color(white)(.) "unit of distance per second}}$

#### Explanation:

Consider x-axis and y-axis as defining the horizontal plane and z-axis as defining the vertical element of 3-space

$\textcolor{b l u e}{\text{Method}}$

Find true length projected on to xy-plane.
Use this and the z-axis value with$\textcolor{b r o w n}{\text{ Pythagoras }}$to find the true length of the vector.

Convert this to speed by dividing by 4 seconds
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given:
$\left({x}_{1} , {y}_{1} , {z}_{1}\right) \to \left(9 , \left(- 6\right) , 1\right)$

$\left({x}_{2} , {y}_{2} , {z}_{2}\right) \to \left(\left(- 1\right) , 5 , \left(- 1\right)\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider proposed method as algebra}}$

$\text{Step 1 xy-plane length} \to \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$\text{Step 2 xy-plane length and z-axis} \textcolor{b l u e}{\to \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Determining magnitude of the vector}}$

let $L$ be magnitude of the vector

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

$L - \sqrt{{\left(- 1 - 9\right)}^{2} + {\left(5 - \left(- 6\right)\right)}^{2} + {\left(- 1 - 1\right)}^{2}}$

$\textcolor{b l u e}{L = \sqrt{100 + 121 + 4} = 15}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine speed}}$

"speed " = ("distance")/("time")

Let the unit of distance be $d$
Let the unit of time be $s$

Given: time$= 4 s$
Calculated distance $L = 15$

$\textcolor{g r e e n}{\implies \text{speed "=15/4color(white)(.) "unit of distance per second}}$