Physics
posted by KGarcia on .
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.
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?

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.