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

Question

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

Amelia Adams · Accepted Answer

#v=sqrt 6" ""unit"/s# #P_1(8,4,1) " "P_2(6,0,2)# #P_"1x"=8" "P_"2x"=6" "Delta P_x=6-8=-2# #P_"1y"=4" "P_"2y"=0" "Delta P_y=0-4=-4# #P_"1z"=1" "P_"2z"=2" "Delta P_ z=2-1=2# #" distance between the point of " P_1" and " P_2" is:"# #Delta x=sqrt((Delta P_x)^2+(Delta P_y)^2+(Delta P_z)^2)# #Delta x=sqrt((-2)^2+(-4)^2+2^2)=sqrt (4+16+4)=sqrt24# #v=(Delta x)/t# #v=sqrt 24/2# #v=sqrt (4*6)/2# #v=(cancel(2)*sqrt6)/cancel(2)# #v=sqrt 6" ""unit"/s#

Layla Ahmed · Answer

undefined To find the speed of the object, we first need to calculate the distance traveled. Using the formula for distance in three-dimensional space, we have:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
$$

Substituting the given coordinates, we get:

$$
\text{Distance} = \sqrt{(6 - 8)^2 + (0 - 4)^2 + (2 - 1)^2} = \sqrt{(-2)^2 + (-4)^2 + 1^2} = \sqrt{4 + 16 + 1} = \sqrt{21}
$$

Now, to find the speed, we use the formula:

$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
$$

Substituting the given time of 2 seconds, we have:

$$
\text{Speed} = \frac{\sqrt{21}}{2} \approx 1.443 \, \text{units per second}
$$

So, the speed of the object is approximately 1.443 units per second.