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April 16, 2014

April 16, 2014

Posted by **KGarcia** on Monday, January 16, 2012 at 10:54pm.

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.

Answer in units of km

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 -
**Henry**, Wednesday, January 18, 2012 at 4:30pmd = 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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