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An athlete leaves one end of a pool of length $$L$$ at $$t = 0$$ and arrives at the other end at time $$t_{1}$$. She swims back and arrives at the starting position at time $$t_{2}$$. If she is swimming initially in the positive $$x$$ direction, determine her average velocities symbolically in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip. (d) What is her average speed for the round trip.

Solution
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We use the definition of average velocity
(a) $$V_{1,x,ave}=\frac{(\Delta x)_{1}}{(\Delta t)_{1}}=\frac{L-0}{t_{1}}=+L/t_{1}$$
(b) $$V_{2,x,ave}=\frac{(\Delta x)_{2}}{(\Delta t)_{2}}=\frac{0-L}{t_{2}}=-L/t_{2}$$
(c) To find the average velocity for the round trip, we add the displacement and time for each of the two halves of the swim
$$V_{x,ave,total}=\frac{(\Delta x)_{total}}{(\Delta t)_{total}}=\frac{(\Delta x)_{1}+(\Delta x)_{2}}{t_{1}+t_{2}}=\frac{+L-L}{t_{1}+t_{2}}=\frac{0}{t_{1}+t_{2}}=0$$
(d) The average speed of the round trip is the total distance the athlete travels divided by the total time for the trip:
$$V_{ave,trip}$$ = total distance travelled / $$(\Delta t)_{total}$$
= $$\frac{|(\Delta x)_{1}|+|(\Delta x)_{2}|}{t_{1}+t_{2}}$$
= $$\frac{|+L|+|-L|}{t_{1}+t_{2}}=\frac{2L}{t_{1}+t_{2}}$$

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