Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Problem 3: The bacterium E.coli is a single cell organism y (am) that lives in t

ID: 1658489 • Letter: P

Question

Problem 3: The bacterium E.coli is a single cell organism y (am) 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 a20 constant speed of 20m/s. Figure shows one such trajectory as the bacteria moves from point A to point E 40 T ID 20 40 6080 100 -20 A. What are the magnitude and direction of the bacterium's net displacement? (where does it start and where does i end up? Ans 45um,,27) B. Write each of the four displacements in vector notation FIGURE P3.40 Write their magnitude and directions. C. Find the time taken for each leg and the total time taken. (Ans. Total time - 8.65 s) D. What are the magnitude and direction of the average velocity for the entire trip Some modifications: How far and in what direction would the bacteria have to travel if it got bored and decided to return to the starting point from point D?

Explanation / Answer

A)

Along Y-direction

Yi = initial position = 0

Yf = Final position = - 20

Y = net displacement along Y-direction = Yf - Yi = - 20 - 0 = - 20

Along X-direction

Xi = initial position = 0

Xf = Final position = 40

X = net displacement along X-direction = Xf - Xi = 40 - 0 = 40

Net displacement is given as

D = sqrt(X2 + Y2) = sqrt((40)2 + (- 20)2) = 45 um ,

direction : theta = tan-1(20/40) = 27 deg below X-axis

b)

AB= 50 i + 10 j

magntitude : sqrt(502 + 102) = 51

direction:  tan-1(10/50) = 11.3 deg above X-axis

BC = 0 i + 10 j

magntitude : sqrt(02 + 102) = 10

direction:  tan-1(10/0) = 90 deg above X-axis

CD= 40 i + 10 j

magntitude : sqrt(402 + 102) = 41

direction:  tan-1(10/40) = 14 deg above X-axis

DE = - 50 i - 50 j

magntitude : sqrt((- 50)2 + (- 50)2) = 71

direction: 180 + tan-1(50/50) = 225 deg counterclockwise from

C)

tAB = 51/20 = 2.55 sec

tBC = 10/20 = 0.5 sec

tCD = 41/20 = 2.05 sec

tDE = 71/20 = 3.55 sec

t = total time = tAB + tBC + tCD + tDE =2.55 + 0.5 + 2.05 + 3.55 = 8.65 sec

D)

average velocity = displacement /time = 45/8.65 = 5.20

direction 27 deg below X-axis

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote