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## Q. 2.1

Swim competition

Now let’s apply our definition of average velocity to a swimming competition. During one heat of a swim meet, a swimmer performs the crawl stroke in a pool 50.0 m long, as shown in Figure 2.3. She swims a length at racing speed, taking 24.0 s to cover the length of the pool. She then takes twice that time to swim casually back to her starting point. Find (a) her average velocity for each length and (b) her average velocity for the entire swim. ## Verified Solution

SET UP As shown in Figure 2.4, we choose a coordinate system with the origin at the starting point (often a convenient choice) and x increasing to the right. We add the information given to the diagram we sketch for the problem.

SOLVE Part (a): For the first length, we have $x_1 = 0, x_2 = 50.0 m, t_1 = 0, and t_2 = 24.0 s.$ Using the definition of average velocity given by Equation 2.1, we find that

$\mathrm{\upsilon _{a,v,x}=\frac{x_2-x_1}{t_2-t_1}=\frac{50.0 m-0m}{24.0s-0s}=2.08 m/s. }$

For the return trip, we have $x_1 = 50.0 m, x_2 = 0, t_1 = 24.0 s, and t_2 = 24.0 s + 48.0 s = 72.0 s.$ Using the definition of average velocity again, we obtain

$\mathrm{\upsilon _{a,v,x}=\frac{x_2-x_1}{t_2-t_1}=\frac{50.0 m-0m}{24.0s-0s}=1.04 m/s. }$

Part (b): The starting and finishing points are the same: $x_1 = x_2 = 0.$ Because the total displacement for a round-trip is zero, the average velocity is zero!

REFLECT The average x component of velocity in part (b) is negative because $x_2$ lies to the left of $x_1$ on the axis and x is decreasing during this part of the trip. Thus, the average velocity vector points to the right for part (a) and to the left for part (b). The average velocity for the total out-and-back trip is zero because the total displacement is zero.

Practice Problem: If the swimmer could cross the English Channel (32 km) maintaining the same average velocity as for the first 50 m in the pool, how long would it take? (Actual times are around 15 hours.) Answer: 4.3 h. 