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

1 Answer
Jan 19, 2016

Answer:

#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
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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")#