A runner is jogging in a straight line at a steady vr= 5 km/hr. When the runner is L= 7.5 km from the finish line, a bird begins flying straight from the runner to the finish line at vb= 20 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.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.

Subtract from L the distance the runner covers when the bird reaches the finish line, to get the length X, then find the time, t₂, it takes the runner and the bird to cover the distance, X, then calculate the distance the bird flies in the time t

Method 2:

Note that the sum of the distance travelled by the bird and the runner when they meet (second time) is equal to twice the distance the distance L

Solution;

The time it takes the bird to reach the finish line, t₁ = 7.5 km/(20 km/hr) = 0.375 hr.

The distance the runner runs in time t₁, d = 5 km/hr × 0.375 hr = 1.875 km

The distance the runner and the bird combined travel before they meet, X, is given as follows;

X = L - d

∴ X = 7.5 km - 1.875 km = 5.625 km

We get; X = ([tex]\mathbf{v_r}[/tex] + [tex]\mathbf{v_b}[/tex]) × t

Which gives;

5.625 = (5 + 20) × t

t = 5.625/(5 + 20) = 5.625/(25) = 0.225

The time it takes for the bird and the runner to meet, t = 0.225 hour

The distance the bird travels before reaching the runner, d₂ = [tex]v_b[/tex] × t

∴ d₂ = 20 km/hr × 0.225 hr = 4.5 km

The cumulative distance the bird travels, [tex]d_t[/tex] = L + d₂

∴ [tex]d_t[/tex] = 7.5 km + 4.5 km = 12 km

The cumulative distance the bird travels, by the time it reaches the runner [tex]d_t[/tex] = 12 km

Alternatively, from the diagram, we have;

[tex]d_b[/tex] = d₁ + d₃

[tex]d_r[/tex] = d₂

d₁ + d₃ + d₂ = 2·L

∴ [tex]d_r[/tex] + [tex]d_b[/tex] = 2·L

Where;

[tex]d_r[/tex] = The distance the runner runs before meeting with the bird

[tex]d_b[/tex] = The distance the bird travels before meeting with the runner

[tex]d_b[/tex] = 4dr (The bird is four times as fast as the runner)

∴ [tex]d_r[/tex] + [tex]d_b[/tex] = 2 × 7.5 = 15

[tex]d_r[/tex] + 4·[tex]d_r[/tex] = 5·[tex]d_r[/tex] = 15

[tex]d_r[/tex] = 15/5 = 3

[tex]d_b[/tex] = 4 × [tex]d_r[/tex]

[tex]d_b[/tex] = 4 × 3 = 12

The distance the bird travels before meeting with the runner, [tex]d_b[/tex] = 12 km

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