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29. Jan 2023•0 gefällt mir•2 views

29. Jan 2023•0 gefällt mir•2 views

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Math 401 Velocity and Net Change Problem 2 (Falling Body.) A Skydiver jumps out of a plane falling(in m/s) at a velocity v(t)-9.8t. After 10 seconds her chute deploys and she instantly slows to a constant speed of 10m/s until she lands on the ground. a. How far does she fall in 30 seconds? b. If she jumped from 3km, when does she land? Solution a) Initial velocity, u = 0 Acceleration in the first 10 s, a = g = 9.8 m/s 2 Using th formula, s = ut + 1/2 gt 2 , where s is the distance, s = 0 + 0.5 x 9.8 x 10 x 10 = 490 m Distance travelled in the next 20 s is, distance = speed x time = 10 x 20 = 200 m Total distance travelled in 30 s = 490 + 200 = 690 m b) Time taken to reach first 490 m is 10 s [From previous part] Time taken to reach next 3000 - 490 = 2510 is t = 2510 / 10 = 251 s Total time taken = 10 + 251 = 261 s .

- 1. Math 401 Velocity and Net Change Problem 2 (Falling Body.) A Skydiver jumps out of a plane falling(in m/s) at a velocity v(t)-9.8t. After 10 seconds her chute deploys and she instantly slows to a constant speed of 10m/s until she lands on the ground. a. How far does she fall in 30 seconds? b. If she jumped from 3km, when does she land? Solution a) Initial velocity, u = 0 Acceleration in the first 10 s, a = g = 9.8 m/s 2 Using th formula, s = ut + 1/2 gt 2 , where s is the distance, s = 0 + 0.5 x 9.8 x 10 x 10 = 490 m Distance travelled in the next 20 s is, distance = speed x time = 10 x 20 = 200 m Total distance travelled in 30 s = 490 + 200 = 690 m b) Time taken to reach first 490 m is 10 s [From previous part] Time taken to reach next 3000 - 490 = 2510 is t = 2510 / 10 = 251 s Total time taken = 10 + 251 = 261 s