Question

001 (part 1 of 2) 10.0 points a runner is jogging in a straight line at a steady $v_r = 8$ km/hr. when the runner is $l = 5.3$ km from the finish line, a bird begins flying straight from the runner to the finish line at $v_b = 32$ km/hr (4 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 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 for bird to reach finish line

The distance to the finish line is \( L = 5.3 \) km and the bird's speed is \( v_b = 32 \) km/hr. Using \( t=\frac{d}{v} \), time \( t_1=\frac{L}{v_b}=\frac{5.3}{32} \) hr. In this time, the runner moves \( d_r1 = v_r \times t_1 = 8\times\frac{5.3}{32}=\frac{5.3}{4}=1.325 \) km. So the distance between runner and finish line when bird reaches finish line is \( L - d_r1 = 5.3 - 1.325 = 3.975 \) km? Wait, no, actually, when the bird reaches the finish line, the time taken is \( t_1=\frac{L}{v_b}=\frac{5.3}{32} \) hr. But wait, another approach: The total time until the runner reaches the finish line is \( T=\frac{L}{v_r}=\frac{5.3}{8} \) hr. Since the bird is flying all the time until the runner reaches the finish line (because when the bird turns back, it will meet the runner before the runner finishes? Wait, no, actually, the bird flies to the finish line, then back to the runner, but the total time the bird is flying is the time until the runner reaches the finish line? Wait, let's check.

Wait, the runner's speed is \( v_r = 8 \) km/hr, distance to finish is \( L = 5.3 \) km. So time for runner to reach finish is \( T=\frac{L}{v_r}=\frac{5.3}{8} \) hr. The bird is flying at \( v_b = 32 \) km/hr for the entire time until the runner reaches the finish line? Wait, no, because when the bird flies to the finish line, then turns back, but does the runner reach the finish line before the bird comes back? Wait, let's calculate the time for bird to go to finish line: \( t_1=\frac{5.3}{32} \) hr. In that time, runner moves \( 8\times\frac{5.3}{32}=1.325 \) km, so distance between runner and finish is \( 5.3 - 1.325 = 3.975 \) km. Now, the bird turns back and flies towards the runner. The relative speed of bird and runner is \( v_b + v_r = 32 + 8 = 40 \) km/hr. The distance between them is 3.975 km, so time to meet is \( t_2=\frac{3.975}{40} \) hr. In that time, bird flies \( 32\times t_2 = 32\times\frac{3.975}{40}=3.18 \) km. Then, after meeting, does the runner still have time to run? Wait, no, maybe the first approach is wrong. Wait, actually, the total time the bird is flying is the time until the runner reaches the finish line. Wait, let's check: runner's time to finish is \( T=\frac{5.3}{8}=0.6625 \) hr. Bird's speed is 32 km/hr, so distance bird flies is \( v_b \times T = 32\times0.6625 = 21.2 \) km? Wait, that can't be, because first the bird flies 5.3 km to finish line, which takes \( 5.3/32 = 0.165625 \) hr. In that time, runner runs \( 8\times0.165625 = 1.325 \) km, so distance between runner and finish is \( 5.3 - 1.325 = 3.975 \) km. Now, bird and runner are moving towards each other: bird speed 32, runner speed 8, so relative speed 40. Time to meet: \( 3.975 / 40 = 0.099375 \) hr. In that time, bird flies \( 32\times0.099375 = 3.18 \) km. Now, after meeting, how far is the runner from finish? Runner has run \( 1.325 + 8\times0.099375 = 1.325 + 0.795 = 2.12 \) km, so distance to finish is \( 5.3 - 2.12 = 3.18 \) km? Wait, no, that's the same as the bird's distance? Wait, no, maybe I messed up. Wait, 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. Because when the bird flies to the finish line and back, but the runner is moving towards the finish line, so the total time until the runner reaches the finish line is \( T = L / v_r \). In that time, the bird is flying at speed \( v_b \), so distance is \( v_b \times T \). Let's verify:

\( T = 5.3 / 8 = 0.6625 \) hr.

Bird's distance: \( 32 \times 0.6625 = 21.2 \) km.

Wait, but when the bird first flies to the finish line, that's 5.3 km, time 5.3/32 = 0.165625 hr. Then, the runner has run 8*0.165625 = 1.325 km, so distance between runner and finish is 5.3 - 1.325 = 3.975 km. Now, bird and runner move towards each other: relative speed 32 + 8 = 40 km/hr. Time to meet: 3.975 / 40 = 0.099375 hr. In that time, bird flies 32*0.099375 = 3.18 km. Now, after meeting, the runner has run 1.325 + 8*0.099375 = 2.12 km, so distance to finish is 5.3 - 2.12 = 3.18 km? Wait, no, that's the same as the bird's distance? Wait, no, now the bird turns back again? Wait, no, this is an infinite series, but the sum of the infinite series is equal to \( v_b \times T \), where \( T \) is the time for the runner to reach the finish line. Let's calculate the sum:

First leg: distance \( d_1 = 5.3 \) km, time \( t_1 = 5.3 / 32 \).

Second leg: distance \( d_2 = 32 \times t_2 \), where \( t_2 = (5.3 - 8t_1) / (32 + 8) \).

But \( 8t_1 = 8*(5.3/32) = 5.3/4 = 1.325 \), so \( 5.3 - 1.325 = 3.975 \), \( t_2 = 3.975 / 40 = 0.099375 \), \( d_2 = 32*0.099375 = 3.18 \).

Third leg: distance \( d_3 = 32 \times t_3 \), where \( t_3 = (3.975 - 8t_2) / (32 + 8) \). \( 8t_2 = 0.795 \), \( 3.975 - 0.795 = 3.18 \), \( t_3 = 3.18 / 40 = 0.0795 \), \( d_3 = 32*0.0795 = 2.544 \). Wait, this is getting smaller, but the sum of the series is \( d_1 + d_2 + d_3 + ... \).

The first term \( a = 5.3 \), common ratio \( r = (v_b - v_r)/(v_b + v_r) = (32 - 8)/(32 + 8) = 24/40 = 0.6 \). Wait, no, the ratio of the distances: \( d_2 = d_1 \times (v_b - v_r)/(v_b + v_r) \)? Wait, no, let's see:

After first leg, distance between runner and finish is \( L - v_r t_1 = L - v_r (L / v_b) = L (1 - v_r / v_b) \).

Then, the bird and runner move towards each other, so the distance between them is \( L (1 - v_r / v_b) \), and their relative speed is \( v_b + v_r \), so time to meet is \( t_2 = [L (1 - v_r / v_b)] / (v_b + v_r) \).

Distance bird flies in \( t_2 \) is \( v_b t_2 = v_b [L (1 - v_r / v_b)] / (v_b + v_r) = L (v_b - v_r) / (v_b + v_r) \).

Then, after meeting, the distance between runner and finish is \( L (1 - v_r / v_b) - v_r t_2 = L (1 - v_r / v_b) - v_r [L (1 - v_r / v_b)] / (v_b + v_r) = L (1 - v_r / v_b) (1 - v_r / (v_b + v_r)) = L (1 - v_r / v_b) (v_b / (v_b + v_r)) = L (v_b - v_r) / (v_b + v_r) \).

Wait, this is the same as the distance the bird flew in \( t_2 \). Then, the next leg, the bird flies to the finish line again? No, wait, when they meet, the bird turns back, so the distance between the new meeting point and the finish line is \( L (v_b - v_r) / (v_b + v_r) \), and the bird flies that distance to the finish line, time \( t_3 = [L (v_b - v_r) / (v_b + v_r)] / v_b = L (v_b - v_r) / [v_b (v_b + v_r)] \).

Then, the runner moves \( v_r t_3 = L v_r (v_b - v_r) / [v_b (v_b + v_r)] \), so the distance between runner and finish is \( L (v_b - v_r) / (v_b + v_r) - L v_r (v_b - v_r) / [v_b (v_b + v_r)] = L (v_b - v_r) (v_b - v_r) / [v_b (v_b + v_r)] = L (v_b - v_r)^2 / [v_b (v_b + v_r)] \).

Then, the bird turns back and flies towards the runner, distance \( d_3 = v_b \times t_4 \), where \( t_4 = [L (v_b - v_r)^2 / (v_b (v_b + v_r))] / (v_b + v_r) = L (v_b - v_r)^2 / [v_b (v_b + v_r)^2] \), so \( d_3 = v_b \times t_4 = L (v_b - v_r)^2 / (v_b + v_r)^2 \).

So the total distance is \( d_1 + d_2 + d_3 + ... = L + 2 \times L (v_b - v_r)/(v_b + v_r) + 2 \times L (v_b - v_r)^2/(v_b + v_r)^2 + ... \). Wait, no, the first leg is \( L \), then the bird flies back and forth, so the series is \( L + 2 \times [L (v_b - v_r)/(v_b + v_r) + L (v_b - v_r)^2/(v_b + v_r)^2 + ...] \).

The sum of the geometric series \( S = a / (1 - r) \), where \( a = L (v_b - v_r)/(v_b + v_r) \), \( r = (v_b - v_r)/(v_b + v_r) \).

So the sum of the back-and-forth legs is \( 2 \times [a / (1 - r)] = 2 \times [L (v_b - v_r)/(v_b + v_r) / (1 - (v_b - v_r)/(v_b + v_r))] = 2 \times [L (v_b - v_r)/(v_b + v_r) / (2 v_r / (v_b + v_r))] = 2 \times [L (v_b - v_r) / (2 v_r)] = L (v_b - v_r) / v_r \).

Then total distance is \( L + L (v_b - v_r) / v_r = L (1 + (v_b - v_r)/v_r) = L (v_b / v_r) \).

Ah! So that's the key. The total distance the bird flies is \( L \times (v_b / v_r) \). Because \( v_b = 4 v_r \), so \( v_b / v_r = 4 \), so distance is \( 5.3 \times 4 = 21.2 \) km.

Let's check with the first approach: runner's time to finish is \( T = L / v_r = 5.3 / 8 = 0.6625 \) hr. Bird's speed is 32 km/hr, so distance is \( 32 \times 0.6625 = 21.2 \) km. Yep, that matches. So the infinite series sum simplifies to \( v_b \times T \), where \( T \) is the time for the runner to reach the finish line, because the bird is flying at constant speed for the entire time until the runner finishes.

Step2: Calculate bird's distance

Using the formula \( d = v_b \times T \), where \( T = L / v_r \), so \( d = v_b \times (L / v_r) \).

Substitute \( v_b = 32 \) km/hr, \( L = 5.3 \) km, \( v_r = 8 \) km/hr:

\( d = 32 \times (5.3 / 8) = (32 / 8) \times 5.3 = 4 \times 5.3 = 21.2 \) km.

Answer:

21.2