Question

In: Physics

What is the magnitude of the bacterium's average velocity for the entire trip?

The bacterium Escherichia coli (or E. coli) is a single-celled organism that lives in the gut of healthy humans and animals. When grown in a uniform medium rich in salts and amino acids, these bacteria swim along zig-zag paths at a constant speed of 20 um/s. The figure shows the trajectory of an E. coli as it moves from point A to point E. Each segment of the motion can be identified by two letters, such as segment BC. 

What is the magnitude of the bacterium's average velocity for the entire trip?

Solutions

Expert Solution

a )IN 1st segment

Vx = 20 μm/s cos(tan inverse (1/5)) = 19.61μm/s
V y = 3.922um/s
In 2nd Segment

Vx=0
Vy = 20um/s
Segment 3 : Vx = 20 μm/s cos(tan inverse (1/4)) = 19.4μm/s
Vy = 4.85um/s
segment 4 : Vx = -20 μm/s cos(45) = -14.14μm/s
V y = -14.14um/s

b)Distance travelled = 172.93μm
displement = 44.72m


c) average velocity = displement/time = 44.72m/8.64sec = 5.17μm/sec
Average speed = distance /time = 172.93/8.64 = 20μm/sec

genius_generous answered 2 years ago
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