Physics Challenge Question 2 Solutions | Boston University - Edubirdie

University:
Boston University
Course:
CAS PY 211 | General Physics I
Academic year:

2024
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433

Pages:

1
Author:

Deja Hood

Physics Challenge Question 2: Solutions Question 1: Time: 0.0 sec Distance: 0.0 ft Velocity: 1.0 sec 2.0 sec 3.0 sec 4.0 sec 5.0 sec 6.0 sec 5.0 ft 15 ft 25 ft 35 ft 55 ft 60 ft 5.0 ft/s 10 ft/s 10 ft/s 10 ft/s 20 ft/s 5.0 ft/s Here, I just calculate the displacement and divide it by the time interval (each of those is 1 sec). For example, the second interval is, with 2 significant figures, 15 ft − 5.0 ft 10 ft v= = = 10 fts 2.0 s − 1.0 s 1.0 s Question 2: The average velocity is (5.0 + 10 + 10 + 10 + 20 + 5.0) v average = 6 ft s = 10 ft s Question 3: I can use the first and last measurement to calculate the average velocity over the whole stretch. This is no coincidence – what happens in between doesn’t matter for the average velocity! 60 ft − 0 ft v average = = 10 fts 6.0 s − 0.0 s Question 4: 1 ft = 0.3048 m, so v average = 10 fts ⋅ 0.3048 mft = 3.0 ms

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