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 ( v_b ), ( v_r ), ( l )

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.

002 (part 2 of 2) 10.0 points
after this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. the bird repeats the back and forth trips until the runner reaches the finish line.

how far does the bird travel from the beginning (including the distance traveled to the first encounter)?

answer in units of km.

Explanation:

Part 1:

Step1: Find time for bird to reach finish

The distance to the finish line is \( L = 5.8 \) km, and the bird's speed is \( v_b = 18 \) km/hr. Using \( t = \frac{d}{v} \), the time for the bird to reach the finish line is \( t_1=\frac{L}{v_b}=\frac{5.8}{18} \) hr. But wait, actually, we can also think about the runner's time to finish. The runner's speed is \( v_r = 6 \) km/hr, so the total time the runner takes to finish is \( T=\frac{L}{v_r}=\frac{5.8}{6} \) hr. Since the bird is flying until the runner finishes (in part 1, wait no, part 1 is the first trip and back? Wait no, part 1: when the bird starts, the runner is \( L = 5.8 \) km from finish. Bird flies to finish: time \( t_1=\frac{5.8}{18} \) hr. In that time, the runner moves \( d_r = v_r \times t_1 = 6\times\frac{5.8}{18}=\frac{5.8}{3} \) km. So the distance between runner and finish when bird reaches finish is \( L - d_r = 5.8 - \frac{5.8}{3}=\frac{2\times5.8}{3} \) km. Now the bird and runner are moving towards each other: relative speed \( v_{rel}=v_b + v_r = 18 + 6 = 24 \) km/hr. Time to meet: \( t_2=\frac{\frac{2\times5.8}{3}}{24}=\frac{5.8}{36} \) hr. Total time for bird in part 1: \( t = t_1 + t_2=\frac{5.8}{18}+\frac{5.8}{36}=\frac{11.6 + 5.8}{36}=\frac{17.4}{36}=\frac{5.8}{12} \) hr? Wait, no, that's complicated. Wait, actually, a simpler way: the total time the bird is flying until the first encounter? Wait no, part 1: "What cumulative distance does the bird travel?" Wait, maybe I misread. Wait, the first part: when the runner is 5.8 km from finish, bird starts flying to finish, then back to runner. Wait, but actually, the key insight is that the total time the bird is flying is the time it takes for the runner to reach the finish line? No, in part 1, it's the first trip to finish and back to runner. Wait, no, let's re-express.

Wait, the runner's speed is 6 km/hr, distance to finish is 5.8 km. The bird's speed is 18 km/hr.

First, time for bird to reach finish: \( t_1 = \frac{5.8}{18} \) hr. In that time, runner moves \( 6 \times \frac{5.8}{18} = \frac{5.8}{3} \) km. So distance between runner and finish when bird reaches finish: \( 5.8 - \frac{5.8}{3} = \frac{2 \times 5.8}{3} \) km. Now, bird and runner are moving towards each other: bird speed 18, runner speed 6, so relative speed 24 km/hr. Time to meet: \( t_2 = \frac{\frac{2 \times 5.8}{3}}{24} = \frac{5.8}{36} \) hr. Total time bird flies: \( t_1 + t_2 = \frac{5.8}{18} + \frac{5.8}{36} = \frac{11.6 + 5.8}{36} = \frac{17.4}{36} = \frac{5.8}{12} \) hr? Wait, no, that can't be. Wait, maybe there's a better way. Wait, the total distance the bird travels is \( v_b \times t \), where \( t \) is the time until the first encounter. But actually, the runner's time to finish is \( T = \frac{5.8}{6} \) hr. Wait, no, in part 1, it's the first trip to finish and back. Wait, maybe I made a mistake. Let's calculate the time when the bird reaches the finish: \( t_1 = 5.8 / 18 \) hours. In that time, runner has run \( 6 * (5.8 / 18) = 5.8 / 3 \) km, so distance from runner to finish is \( 5.8 - 5.8/3 = 11.6/3 \) km. Now, bird and runner are moving towards each other: bird speed 18, runner speed 6, so combined speed 24 km/hr. Time to meet: \( (11.6/3) / 24 = 11.6 / 72 = 5.8 / 36 \) hours. Total time bird is flying: \( 5.8/18 + 5.8/36 = (11.6 + 5.8)/36 = 17.4/36 = 0.4833... \) hours. Then distance bird travels is \( 18 * 0.4833... = 8.7 \) km? Wait, but let's check with another approach. The runner's speed is 6, bird's speed is 18, so the ratio of speeds is 1:3. When the bird reaches the finish, the runner has covered 1/3 of…

Step1: Find total time runner takes to finish

The runner's speed is \( v_r = 6 \) km/hr, distance to finish is \( L = 5.8 \) km. So total time \( T = \frac{L}{v_r} = \frac{5.8}{6} \) hr.

Step2: Calculate bird's total distance

The bird is flying at speed \( v_b = 18 \) km/hr for time \( T \). So distance \( d = v_b * T = 18 * \frac{5.8}{6} \).

Simplify: \( 18/6 = 3 \), so \( d = 3 * 5.8 = 17.4 \) km.

Answer:

(Part 1):
\( 8.7 \) km