question archive The bacterium E
The bacterium E
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The bacterium E. coli is a single-cell organism that lives in the gut of healthy animals, including humans. When grown in a uniform medium in the laboratory, these bacteria swim along zigzag paths at a constant speed of 24 . The following figure shows the trajectory of anE. coli as it moves from point A to point E.(Figure 1)
Part A
What is the magnitude of the bacterium's average velocity for the entire trip?
Part B
What is the direction of the bacterium's average velocity for the entire trip?
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Average Velocity = Net Displacement/Total Time
HereNet Displacement of the Final Position E from the O = sqrt(40^2+20^2)
= 44.721 um
Here
Length AB = sqrt(10^2+50^2)
= 51 um
Time Taken = 51/24
= 2.125 sec
CB Length = 10 um
Time = 10/24
= 0.41667 sec
Length CD = sqrt(10^2+40^2)
= 41.23 umTime = 41.23/24
= 1.718 sec
Length DE = sqrt(50^2 + 50^2)
= 70.71 um
Time = 70.71/24
= 2.946 sec
ThereforeAverage Velocity = 44.721/(2.125+0.41667+1.718+2.946)
= 6.206 um/sec
It Makes an Angle arctan(20/40)
= 26.565 degree Clockwise with X Axis in Fourth Quadrant
