Question

001 (part 1 of 2) 10.0 points
a runner is jogging in a straight line at a steady ( v_r = 6 ) km/hr. when the runner is ( l = 5.8 ) km from the finish line, a bird begins flying straight from the runner to the finish line at ( v_b = 18 ) km/hr (3 times as fast as the runner). when the bird reaches the finish line, it turns around and flies directly back to the runner.
image of runner, bird, finish line with labels
what cumulative distance does the bird travel? even though the bird is a dodo, assume that it occupies only one point in space (a \zero\ length bird), travels in a straight line, and that it can turn without loss of speed.
answer in units of km.

Explanation:

Step1: Find time to finish line

First, find the time it takes for the bird to reach the finish line. The distance to the finish line is \( L = 5.8 \) km, and the bird's speed is \( v_b = 18 \) km/hr. Using the formula \( t = \frac{d}{v} \), the time \( t_1 \) is \( t_1=\frac{L}{v_b}=\frac{5.8}{18} \) hours. But wait, actually, we can also think about the runner's motion. However, a simpler way: the total time until the runner reaches the finish line is \( t_{total}=\frac{L}{v_r}=\frac{5.8}{6} \) hours. Since the bird is flying the entire time the runner is moving until they meet (or the runner finishes), the bird's total time flying is the same as the time it takes for the runner to reach the finish line (because when the runner finishes, the bird will have stopped, or we can calculate the total time the bird is in motion). Wait, let's check: when the bird first flies to the finish line, time \( t_1 = \frac{5.8}{18} \) hr. In that time, the runner moves \( d_r1 = v_r \times t_1 = 6 \times \frac{5.8}{18}=\frac{5.8}{3} \) km. So the distance between the runner and the finish line when the bird reaches the finish line is \( 5.8 - \frac{5.8}{3}=\frac{2\times5.8}{3} \) km. Then the bird and the runner are moving towards each other: relative speed \( v_{rel}=v_r + v_b = 6 + 18 = 24 \) km/hr. Time to meet \( t_2=\frac{\frac{2\times5.8}{3}}{24}=\frac{5.8}{36} \) hr. Total time the bird flies is \( t_1 + t_2=\frac{5.8}{18}+\frac{5.8}{36}=\frac{11.6 + 5.8}{36}=\frac{17.4}{36}=\frac{5.8}{12}=\frac{2.9}{6} \) hr? Wait, no, that's complicated. Wait, actually, the total time the bird is flying is the same as the time it takes for the runner to reach the finish line. Because when the runner reaches the finish line, the bird will have made its trips. Let's verify: runner's time to finish is \( t=\frac{L}{v_r}=\frac{5.8}{6} \) hours. In that time, the bird flies at \( v_b = 18 \) km/hr, so distance is \( d = v_b \times t = 18 \times \frac{5.8}{6} \). Let's calculate that.

Step2: Calculate bird's distance

Simplify \( 18 \times \frac{5.8}{6} \). \( 18/6 = 3 \), so \( 3 \times 5.8 = 17.4 \) km. Wait, that's simpler. Because the bird is flying the entire time the runner is moving towards the finish line. So regardless of the turns, the total time the bird is in motion is the time it takes the runner to reach the finish line. So time \( t = \frac{L}{v_r} = \frac{5.8}{6} \) hours. Then distance for bird is \( v_b \times t = 18 \times \frac{5.8}{6} = 3 \times 5.8 = 17.4 \) km. Wait, but let's check with the first method. First trip: bird flies 5.8 km, time \( 5.8/18 \) hr. Runner moves \( 6*(5.8/18)=5.8/3 \) km, so remaining distance between runner and finish is \( 5.8 - 5.8/3 = 11.6/3 \) km. Now, bird and runner move towards each other: speed bird 18, runner 6, so relative speed 24. Time to meet: \( (11.6/3)/24 = 11.6/(72) = 5.8/36 \) hr. In that time, bird flies \( 18*(5.8/36)=5.8/2 = 2.9 \) km. Then, after meeting, the bird would turn around, but wait, when they meet, has the runner reached the finish line? Wait, no, after the first trip, runner has \( 11.6/3 \) km left, which is about 3.866 km. At 6 km/hr, time to finish from there is \( (11.6/3)/6 = 11.6/18 = 5.8/9 \approx 0.644 \) hr. But the bird and runner meet in \( 5.8/36 \approx 0.161 \) hr, during which runner moves \( 6*(5.8/36)=5.8/6 \approx 0.966 \) km, so runner's total distance from start is \( 5.8/3 + 5.8/6 = 11.6/6 + 5.8/6 = 17.4/6 = 2.9 \) km, so distance to finish is \( 5.8 - 2.9 = 2.9 \) km. Now, the bird has flown 5.8 + 2.9 = 8.7 km? Wait, that's conflicting. Wait, no, I made a mistake. Wait, first trip: bird flies 5.8 km, time \( t1 = 5.8 / 18 \) hr. Runner's distance: \( 6 * t1 = 6*(5.8/18) = 5.8/3 \approx 1.933 \) km. So distance between runner and finish: \( 5.8 - 1.933 = 3.866 \) km. Now, bird turns around, flies back towards runner. Now, runner is moving towards finish at 6 km/hr, bird towards runner at 18 km/hr. So relative speed is 6 + 18 = 24 km/hr. Time to meet: \( t2 = 3.866 / 24 \approx 0.161 \) hr. In t2, bird flies \( 18 * 0.161 \approx 2.9 \) km. Now, runner's distance in t2: \( 6 * 0.161 \approx 0.966 \) km. So total runner distance: \( 1.933 + 0.966 \approx 2.9 \) km, distance to finish: \( 5.8 - 2.9 = 2.9 \) km. Now, bird turns around again, flies to finish: time \( t3 = 2.9 / 18 \approx 0.161 \) hr, distance \( 18 * 0.161 \approx 2.9 \) km. Now runner's distance in t3: \( 6 * 0.161 \approx 0.966 \) km, total runner distance: \( 2.9 + 0.966 \approx 3.866 \) km, distance to finish: \( 5.8 - 3.866 \approx 1.933 \) km. This is an infinite series, but the total time until the runner finishes is \( 5.8 / 6 \approx 0.9667 \) hr. Let's calculate the total distance the bird flies: \( 18 * 0.9667 \approx 17.4 \) km. Ah, so the infinite series sums up to that. So the key insight is that the total time the bird is flying is the time it takes the runner to reach the finish line, because once the runner reaches the finish line, the bird stops (or the problem implies that we can consider the total time the bird is in motion as the time until the runner finishes, because the bird's motion is continuous until the runner reaches the finish line). So using \( t = L / v_r \), then distance \( d = v_b * t = 18 * (5.8 / 6) = 3 * 5.8 = 17.4 \) km.

Answer:

\( \boxed{17.4} \)