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
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
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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?
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Ok, the bird travels 2.6km to the finish line after it leaves the runner. The runner travels 0.52km during this time as you said. So the distance between the runner and bird (at the finish line) is 2.6-0.52=2.08km. The closing speed is 50.4km/h. Ok, the bird travels five times as fast as the runner. This means that it travels (4/5)*2.08 km in the same time that the man travels (1/5)*(2.08) km. This is where they meet. The bird travels 1.66 + 2.6=4.26km in total.
Nope that seems to be incorrect as well...
does it want total distance travelled or net displacement?
Net displacement would be: 0.52+ (1/5)*2.08=0.94km
it does want total distance... sorry my internet shut down for some reason.
hmmm not sure where my error is, but i'll see if I can find it