Explanation: From a velocity vs time graph, the displace- ment of an object is determined by the area under the curve. When the velocity increases linearly from zero to tQ , the displacement will increase pro- portional to time squared. While the velocity remains constant from tQ to t3, the displacement will increase lin- early. Finally, while the velocity decreases lin- early, the displacement will increase propor- tional to negative time squared. From these facts, we can obtain the correct answer . A | | | | | 005(partlof2)10.0points A runner is jogging in a straight line at a steady 11,-: 5 km/hr. When the runner is L: 3.9 km from the finish line, a bird begins ying straight from the runner to the finish line at 113,: 15 km/ hr (3 times as fast as the runner). When the bird reaches the nish line, it turns around and ies directly back to the runner. (i\\\\/ v? . finish r X line What cumulative distance does the bird travel? Even though the bird is a dodo, as- sume that it occupies only one point in space (a "zero" length bird), travels in a straight line, and that it can turn without loss of speed. Correct answer: 5.85 km. Explanation: Let: 1),. =5km/hr, L=3.9km, and vb=3v,.. _-\\ .--.;, l\\\\.\\ (7/ _ 0&3}; ' _.J / :~ k :.-' . \\ V... if \\ ____,I\\, nish g Z \\' line \"*1 dbl 'J The runner travels a distance 1: until the encounter with the bird. In that time, the bird has traveled a distance L + (L 3:) = 2L 1:. The bird travel 3 times as fast as the runner during this time frame, so db=3dr 2La:=3a: 2L=4m 1 $=L and the bird ies a distance of 1 3 3 d=2L-L=-L=- .k b 2 2 2(39 m) = 5.85km . 006(part20f2)10.0points