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
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?
The first piece is right of course, the bird first flies 2.6 km. I also agree that runner covers 0.52 km in that time. Start the clock again at that moment with t = 0. Write down the equations of motion now for runner and bird and solve them carefully. Runner's position, r(t) = -(2.6 - 0.52) + 8.4t Bird's position, b(t) = -42t Now find out for what t: r(t) = b(t) and then use that to calculate the length of the bird's second leg of flight.
Make sense?
i understand what your saying (and sorry my internet connection seems to be troubled) -2.08 + 8.4t = -42t - 8.4t = -8.4 -2.08/-33.6=-33.6t/-33.6 = .06 * Birds velocity (42) = 2.6 doesnt seem correct.
Careful: -2.08 + 8.4t = -42t ... yes. But now... -2.08 = -50.4t => t = 0.04127 hours Hence the distance the bird covers in the second leg is:
oh goodness.... 1.68 + 2.6 = 4.28
I get (distance of 2nd leg) = 0.0413 hours * 42 km/hr = 1.73 km
Yes. Apparently my finger slipped hen typing into the calculator. Thank you again.
Sure
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