# Movement of bacterium in the microscope

Tags:
1. Jan 17, 2014

### negation

1. The problem statement, all variables and given/known data

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

3. 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

2. Jan 17, 2014

### voko

Looks good.

3. Jan 17, 2014

### 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.

4. Jan 17, 2014

### negation

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

5. Jan 19, 2014

### 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)