Movement of bacterium in the microscope

[negation]

Homework Statement

A biologist looking through a microscope sees a bacterium at r1→ = 2.2i + 3.7j -1.2kμm.
After 6.2s, it's at r2→ = 4.6i + 1.9kμm.
a)What is it's average velocity
b) What is its average speed

The Attempt at a Solution

a)
v→= Δr/Δt
v = (r2-r1)/6.2 = (2.4i - 3.7j + 3.1k)μm/6.2s
= (0.387i -0.596j + 0.5k) μm^-1

b)
|v| = Δs/Δt
|v| = SQRT[(2.4)^2i + (-3.7)^2j + (3.1)^2k]/6.2s
= 0.869μms^-1

 

[voko]

Looks good.

 

[lightgrav]

Caution: IF it went from the first location to the second location by a long and winding path,
then the average velocity will be the same but the average speed might have been many times that fast:
so that is a _minimum_ value for the average speed.

 

[negation]

Caution: IF it went from the first location to the second location by a long and winding path,
then the average velocity will be the same but the average speed might have been many times that fast:
so that is a _minimum_ value for the average speed.

If the path was curved, I would have to utilize optimization, am I right?

 

[lightgrav]

if the path had been curved, you would need to find out long that path was
by adding each small segment length, found via pythagoras (or integrating the path if given a function with time)