# Movement of bacterium in the microscope

## 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

Looks good.

lightgrav
Homework Helper
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.

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
Homework Helper
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)