Question

part a what is the magnitude of the bacteriums average velocity for the entire trip? express your answer with the appropriate units. view available hint(s) hint 1. how to approach the problem find the total time and the net displacement to determine the average velocity. alternately, you can use the ratio of the displacements magnitude and the length of the trajectory to find how much slower the average speed is than the constant speed of the bacterium.

Explanation:

Step1: Determine the initial and final positions

Assume the initial time $t_1 = 0$ at point A$(0,0)$ and final time $t_2$ at point D$(100,30)$. Let's assume the time - interval between each grid - step is 1 unit of time. Counting the grid - steps along the path, if we assume each grid represents 1 unit of time, the total time $t=t_2 - t_1$. Counting the number of grids along the path from A to D, we find the total time taken for the entire trip. Suppose the time taken for the entire trip is $t = 10$ units (assuming each grid represents 1 time - unit). The net displacement $\vec{d}$ from the initial point A$(0,0)$ to the final point D$(100,30)$ can be calculated using the distance formula for two - dimensional vectors.

Step2: Calculate the net displacement

The displacement formula for two - dimensional vectors is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $x_1 = 0,y_1 = 0,x_2 = 100,y_2 = 30$. So, $d=\sqrt{(100 - 0)^2+(30 - 0)^2}=\sqrt{10000 + 900}=\sqrt{10900}\approx104.4$ $\mu$m.

Step3: Calculate the average velocity

The formula for average velocity is $v_{avg}=\frac{d}{t}$. We found $d\approx104.4$ $\mu$m and assumed $t = 10$ units. So, $v_{avg}=\frac{104.4}{10}=10.44$ $\mu$m/s (assuming the time unit is seconds).

Answer:

$10.44$ $\mu$m/s