# Physics

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A runner is jogging at a steady 8.4 km/hr.
When the runner is 2.6 km from the finish
line, a bird begins flying from the runner to
the finish line at 42 km/hr (5 times as fast
as the runner). When the bird reaches the
finish line, it turns around and flies back to
the runner.
L
vb
vr
finish
line
How far does the bird travel? Even though
the bird is a dodo, assume that it occupies
only one point in space (a “zero” length bird)
and that it can turn without loss of speed.

This is my take on it :Flight time of bird to finish line FL = 2.6/42 =(approx) .06hr
Distance runner travels during time period = .06(8.4) = .52km
Distance of runner to FL = 2.6-.52 = 2.08km
Net closing speed between runner and bird = 8.4km +42 km = 50.4km
Time for runner and bird to meet 2.6/50.4 = (approx) .05hr
Distance covered by runner (.05)8.4 = (approx) .43 km
Distance covered by bird to runner (.05)(42) = (approx) 2.1km

2.1 + 2.6 = 4.7km (its not correct) :( What am I doing wrong?

• Physics -

d = Vt = 2.6 km.
42t = 2.6,
t = 2.6/42 = 0.062 h. = Time for bird to reach finish line.

d = Vt=8.4 * 0.062 = 0.521 km=Distance
covered by runner as bird traveled to finish line.

Dr = 2.6 - 0.52 = 2.08 km = Runner's distance when bird reached finish line.

dr + db = 2.08 km,
8.4t + 42t = 2.08,
50.4t = 2.08,
t = 2.08/50.4 = 0.04127 h.

db = Vt = 42 * 0.04127 = 1.733 km. =
Dist. traveled by bird while meeting runner.

Db = 2.6 + 1.733 = 4.33 km.=Tot. dist.
traveled by bird.

Even tho your approach was different from mine, I believe it was correct.
But you made one error: 2.6/50.4 shouled be 2.08/50.4.

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