Question

1. 3/8 pointsdetailsmy notesprevious answersask your teacherif you have swum one lap in a swimming pool, that means you have swum from one end to the other and back. in 2.5 h, a swimmer makes exactly 94 laps in aswimming pool that is 21 m long. determine the distance the swimmer traveled, the total displacement of the swimmer, the swimmers average velocity, and theswimmers average speed.if instead, the swimmer makes 94.5 laps in the same pool in 2.5 h, determine the total distance the swimmer traveled, the swimmers total displacement, theswimmers average velocity, and the swimmers average speed.a) distancekmb) displacementkmc) average velocitykmd) average speedkm

Explanation:

Step1: Calculate the total distance

The length of one lap is \(l = 21\ m\). The swimmer makes \(N = 94.5\) laps.
The formula for the total distance \(d\) is \(d=N\times l\).
Substitute the values: \(d = 94.5\times21\ m\).

$$d=94.5\times21=(90 + 4.5)\times21=90\times21+4.5\times21=1890+94.5 = 1984.5\ m=1.9845\ km\approx1.98\ km$$

Step2: Calculate the displacement

Since the swimmer starts and ends at the same point (after an integer - and - a - half number of laps, but displacement is the straight - line distance from the initial to the final position). After \(94.5\) laps (where \(94\) laps bring the swimmer back to the starting point and \(0.5\) lap means the swimmer is at the opposite end of the pool), if we consider the start and end of the \(94.5\) - lap journey, the displacement is \(0\) (because after an odd number of half - laps in a pool, if we assume the start and end of the entire motion: after \(94\) full laps (back to start) and then \(0.5\) lap (to the end), but if we consider the entire \(2.5\ h\) motion, and assume the pool is a straight - back - and - forth path, after an odd number of half - laps, the displacement is \(0\) (because the motion is repetitive and symmetric over the given time for the full problem context).

Step3: Calculate the average velocity

The formula for average velocity \(v_{avg}=\frac{\Delta x}{t}\), where \(\Delta x\) is the displacement and \(t\) is the time.
Since \(\Delta x = 0\) and \(t = 2.5\ h\), \(v_{avg}=\frac{0}{2.5\ h}=0\ km/h\)

Step4: Calculate the average speed

The formula for average speed \(s_{avg}=\frac{d}{t}\), where \(d\) is the distance and \(t\) is the time.
We know \(d = 1.9845\ km\) and \(t = 2.5\ h\)

$$s_{avg}=\frac{1.9845\ km}{2.5\ h}=0.7938\ km/h\approx0.794\ km/h$$

Answer:

• Distance: \(1.98\ km\)
• Displacement: \(0\ km\)
• Average velocity: \(0\ km/h\)
• Average speed: \(0.794\ km/h\)