Question

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

Answer 1

`color(green)("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

`color(blue)("Method")`

Find true length projected on to xy-plane.
Use this and the z-axis value with`color(brown)(" Pythagoras ")`to find the true length of the vector.

Convert this to speed by dividing by 4 seconds
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Given:
`(x_1,y_1,z_1)->(9,(-6),1)`

`(x_2,y_2,z_2)->((-1),5,(-1))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Consider proposed method as algebra")`

`"Step 1 xy-plane length"->sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"Step 2 xy-plane length and z-axis"color(blue)(->sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determining magnitude of the vector")`

let `L` be magnitude of the vector

`L=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)` becomes:

`L-sqrt((-1-9)^2+(5-(-6))^2+(-1-1)^2)`

`color(blue)(L=sqrt(100+121+4)=15)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine speed")`

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

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

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

`color(green)(=>"speed "=15/4color(white)(.) "unit of distance per second")`