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Home Newtonian Physics Acceleration and Free Fall Examples A skydiver  
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A skydiver
The graphs in figure k show the results of a fairly realistic computer simulation of the motion of a skydiver, including the effects of air friction. The x axis has been chosen pointing down, so x is increasing as she falls. Find (a) the skydiver's acceleration at t = 3.0 s, and also (b) at t = 7.0 s. The solution is shown in figure l. I've added tangent lines at the two points in question.
(a) To find the slope of the tangent line, I pick two points on the line (not necessarily on the actual curve): (3.0 s, 28m/s) and (5.0 s, 42 m/s). The slope of the tangent line is (42 m/s  28 m/s)/(5.0 s  3.0 s) = 7.0 m/s^{2}. (b) Two points on this tangent line are (7.0 s, 47 m/s) and (9.0 s, 52 m/s). The slope of the tangent line is (52 m/s  47 m/s)/(9.0 s  7.0 s) = 2.5 m/s^{2}. Physically, what's happening is that at t = 3.0 s, the skydiver is not yet going very fast, so air friction is not yet very strong. She therefore has an acceleration almost as great as g. At t = 7.0 s, she is moving almost twice as fast (about 100 miles per hour), and air friction is extremely strong, resulting in a significant departure from the idealized case of no air friction.


Home Newtonian Physics Acceleration and Free Fall Examples A skydiver 