If the swimmer could cross the English Chanel (32km) maintains the same velocity as for the first 50 m in the pool, how long would it take? (Actual times are around 15 hours) answer 4.3 hPhotos have background info. Solve only the practice problem. 30CHAPTER 2 Motion Along a Straight LineFor the return trip, we have x₁ = 50.0 m, x₂ = 0, t₁ = 24.0 s, and12= 24.0 s + 48.0 s = 72.0 s. Using the definition of average veloc-ity again, we obtain0m - 50.0 m72.0 s 24.0 s= -1.04 m/s.Uav, xx2-x112-41Part (b): The starting and finishing points are the same: x₁ = x₂ = 0.Because the total displacement for a round-trip is zero, the averagevelocity is zero!REFLECT The average x component of velocity in part (b) is negativebecause x₂ lies to the left of x₁ on the axis and x is decreasing duringthis part of the trip. Thus, the average velocity vector points to the rightfor part (a) and to the left for part (b). The average velocity for the totalout-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 inthe pool, how long would it take? (Actual times are around 15 hours.)Answer: 4.3 h.You may have noticed that we haven't used the term speed in our discussion of particlmotion. The terms speed and velocity are often used interchangeably (and erroneouslybut there is a very important distinction between them. The average speed of a particlduring any motion is a scalar quantity equal to the total distance traveled (disregardinthe direction of motion) during a specified time interval, divided by the time interval. Thaverage velocity is a vector quantity, equal to the vector displacement of the particle dvided by the time interval. In Example 2.1, the average velocity during the two laps is zebecause the total displacement is zero. But the average speed is the total distance travel-(100.0 m) divided by the total time (72.0 s): 100.0 m/72.0 s = 1.39 m/s. The speedometer of a car is correctly named; it measures speed, irrespective of the direction of moticTo make a velocity meter, you would need both a speedometer and a compass.In Example 2.1, we could just as well have chosen the origin 2581.0 s) 3,0 s. We denterval (3.0 s); that is,vector quantity whoset of the car is at positime interval from ax); all other com.Hong the x axis. We1x1 to X2 dividedent ofoccurs. Wee value"(2.1)In the preceding example, x, 19 m, xy277 m. f, 1.0 s, and 4.0 s, andEquation 2.1 givesPart a)-86 m/s.In some cases, U is negative, as shown in Figure 2.2. Suppose a race official's truck ismoving toward the origin, with xy = 277 m x₂=19 m. If t₁= 6.0 s and ₂ 16.0 s,then Ax=-258 m, Ar 10.0 s, and the x component of average velocity is -26 m/s.Whenever x is positive and decreasing, or negative and becoming more negative, the objectis moving in the negative x direction and Dav.x is negative.We stress again that average velocity is a vector quantity, and Equation 2.1 definesthe x component of this vector. In this chapter, all vectors have only one nonzero compo-nent, frequently only an x component. We'll often call Ax the displacement and Dav.x theaverage velocity, remembering that these are really the x components of vector quantities that,in straight-line motion, have only x components. In Chapter 3, we will learn how to deal withdisplacement, velocity, and acceleration vectors that have two nonzero components.In Figures 2.1 and 2.2, it doesn't matter whether the car's velocity is or isn't constantduring the time interval Ar. Another dragster may have started from rest, reached a maxi-mum velocity, blown its engine, and then slowed down. To calculate the average velocity,we need only the total displacement A.x = x₂ x₁ and the total time interval At = 1₂ 1₁.Part b)EXAMPLE 2.1 Swim competitionNow let's apply our definition of average velocity to a swimming competition. During one heat of a swimmeet, a swimmer performs the crawl stroke in a pool 50.0 m long, as shown in Figure 2.3. She swimsa length at racing speed, taking 24.0 s to cover thelength of the pool. She then takes twice that time toswim casually back to her starting point. Find (a) heraverage velocity for each length and (b) her averagevelocity for the entire swim.x₁ = 0t₁ = 0Ugy.xSOLUTIONSET UP As shown in Figure 2.4, we choose a coordinate systemwith the origin at the starting point (often a convenient choice) and x+>xVav, xA277 m4.0 sX₂=0t₂ = 72.0 sA FIGURE 2.4 Our sketch for this problem.19 m1.0 sVav, xA FIGURE 2.3x₂ = 50.0 mt₂ = 24.0 sX₁ = 50.0 mt₁ = 24.0 s50.0 m01increasing to the right. We add the information given to the diagram wesketch for the problem.Uav.x =SOLVE Part (a): For the first length, we have x₁ = 0, x₂ = 50.0 m,t₁ = 0, and t₂ = 24.0 s. Using the definition of average velocity givenby Equation 2.1, we find thatx2x112-11Video Tutor Solution50.0 m 0 m24.0 s0s= 2.08 m/s.CONTINUED

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Description: If the swimmer could cross the English Chanel (32km) maintains the same velocity as for the first 50 m in the pool, how long would it take? (Actual times are around 15 hours) answer 4.3 hPhotos have background info. Solve only the practice problem. 3