Vectors and motion in 2D (physics problem)

[natty210]

vectors and motion in 2D (physics 1111 problem!)

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$$\mu$$m. 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.

(graph is attached!)***

a) for each of the four segments in the bacterium's trajectory, calculate the x and y components 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 and for the entire trip?

 

[tiny-tim]

Welcome to PF!

Hi natty210! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[tomcue]

I would approach this problem by first identifying the given information and variables. In this case, we are dealing with motion in 2D, specifically the movement of the bacterium E. coli. The given speed for the bacterium is 20 micrometers per second and the trajectory is represented by a zig-zag path from point A to point E.

a) To calculate the x and y components of the bacterium's velocity, we can use the formula v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken. Since the bacterium is moving at a constant speed, we can assume that the time taken for each segment is the same. Therefore, for segment AB, the velocity components would be 20\mum/s in the x-direction and 0\mum/s in the y-direction. For segment BC, the velocity components would be 10\mum/s in the x-direction and 10\mum/s in the y-direction. Similarly, for segments CD and DE, the velocity components would be 20\mum/s and 0\mum/s respectively in the x-direction, and 0\mum/s and -20\mum/s respectively in the y-direction.

b) The total distance traveled by the bacterium can be calculated by adding up the distances traveled in each segment. In this case, it would be 20\mum + 20\mum + 20\mum + 20\mum = 80\mum. The magnitude of the net displacement can be calculated using the Pythagorean theorem, which would be the square root of the sum of the squares of the x and y displacements. In this case, it would be sqrt((20\mum)^2 + (-20\mum)^2) = 28.28\mum.

c) The magnitude of the bacterium's average velocity can be calculated by dividing the total distance traveled by the total time taken. Since the time taken for each segment is the same, we can use the formula v_avg = d/t. In this case, it would be 80\mum/4s = 20\mum/s. The direction of the average velocity can be determined by considering the net displacement. In this case, the bacterium moved 20\mum in the x-direction and -20\