The bacterium Escherichia coli (or E. coli) is a...

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 $20 \mu \mathrm{m} / \mathrm{s} .$ Figure $\mathrm{P} 3.52$ 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. For each of the four segments in the bacterium's trajectory, calculate the $x$ - and $y$ -components of its displacement and of its velocity. b. Calculate both the total distance traveled and the magnitude of the net displacement for the entire motion. c. What are the magnitude and the direction of the bacterium's average velocity for the entire trip?

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 So here we are given that the bacterium equally is a single -celled organism that lives in the guard of healthy human animals.

00:11 And when grown into a uniform reach in salts and amino acids, bacteria swim along zig -sacept path at a constant speed of 11 milt, 20 micrometers per second, right? and we are given the trajectory of this equally bacteria.

00:27 So let me just draw this simple diagram.

00:31 From origin i just note the points that's enough there is b this the point a and b the coordinate of this b is 2040 right xp f c the coordinates are 6040 right and we have the final okay so there is another d coordinate which is 9010 and the final coordinate is e and this is 10 minus 40 right so it moves along this path a b c d e right so first of all you have to calculate the magnitude of the net okay, so the total distance covered right so total distance is the sum of all these distances right so we calculate first of all we start with calculating a b right now a b is root over of the thing is very pretty simple root of 40 minus zero squared plus here we have 20 minus 0 squared right this is equal to let me see how much we're getting it's a 44 .72 micrometer similarly we have got bc that is pretty simple right so here we have 40 as same so the x coordinate is for 60 minus 20 which is 40 right cd again we have to apply this distance formula so the x is 90 minus 60 also plus here we have 40 minus 10 whole squared 10 minus 40 also you can do the whatever you can do because we have quad squared so anyways that will be positive right this is equal to 42 .4 3 mu meters and the final one d you have to take the distance formula so 90 minus 10 plus 10 plus 40 right now you have to apply plus because there is minus so this gives us 94 3 4 mu am right now the distance d will be some of all of this rate so this will give us let me just calculate and see 44 .72 plus 40 plus 42 .43 plus 94 .34 3 4 .34 right this is the total distance is 2 .70.

03:56 41 .49...

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