The bacterium Escherichia coli (or E. coli) is a single celled organism that liv
ID: 2271676 • Letter: T
Question
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 medium rich in salts and amino acids, these bacteria swim along zig zag paths at a constant speed of 20 micrometers per second. The figure shows the trajectory of 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.
PLEASE ANSWER ALL QUESTIONS THANK YOU!!!!!
A) Calculate the total distance traveled
B) Calculate the magnitude of the net displacement for the entire motion
C) What is the magnitude of the bacterium's average velocity for the entire trip?
(answers A-C should be in micrometers per second)
D) What is the direction of the bacterium's average velocity for the entire trip? (answer should be in degrees)
So far I have calculated the x and y components of the displacement (x, y [x is first then y])
1. AB = 50, 10
2. BC = 0, 10
3. CD = 40, 10
4. DE = -50, -50
And the x and y components of the velocity
1. AB = 20, 3.9
2. BC = 0, 20
3. CD = 19, 4.9
4. DE= -14, -14
Explanation / Answer
a) Total Distance = AB + BC + CD + DE = sqrt(50^2 + 10^2) + (20-10) + sqrt(40^2 + 10^2) + sqrt(50^2 + 50^2)
Distance = 172.93 micro-meter
b) Displacement = AE = sqrt(40^2 + -20^2) = 44.72 um
Angle = tan^-1(20/40) = 26.56 degree ; clockwise from +ve x axis
c) Time = Distance / speed = 172.93 / 20 = 8.6465 sec
So, Avg velocity = displacement / time = 44.72/8.6465 = 5.172 um/s
d) Same as that of displacement = 26.56 degree ; clockwise from +ve x axis
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