Exam1Sol - AB A 10.0m/s 3.83m Introductory Physics 3.00m...

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3.00m 20.0 o A B 3.83m θ 3.53m 10.0m/s θ Introductory Physics I PHYS 2211-160 Exam #1 Chapters 1-4 Name_____________________________ Show all your work on these sheets, and make sure to box your answers! For problems 1 – 4 the position of an object is described by the equation x (t) = [3m + (4m/s) t + (2m/s 3 ) t 3 ] x . 1. (10 points) What is the displacement vector of the object from t 1 = 2s to t 2 = 4s ? We have that x = x 2 x 1 , and with x 2 = x (4s) = [3m + (4m/s) (4s) + (2m/s 3 ) (4s) 3 ] x = 147m x , and x 1 = x (2s) = [3m + (4m/s) (2s) + (2m/s 3 ) (2s) 3 ] x = 27m x , so x = 147m x - 27m x = x = 120m x . 2. (10 points) What is the average velocity vector of the object from t 1 = 2s to t 2 = 4s ? We have that v av = x / t = 120m x / (4s – 2s) = v av = (60m/s) x . 3. (10 points) What is the velocity vector as a function of time for the object? We have that v (t) = d x / dt = (d / dt) [3m + (4m/s) t + (2m/s 3 ) t 3 ] x = [0 + (4m/s) + (6m/s 3 ) t 2 ] x = v (t) = [(4m/s) + (6m/s 3 ) t 2 ] x . 4. (10 points) What is the acceleration vector as a function of time for the object? We have that
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This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.

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Exam1Sol - AB A 10.0m/s 3.83m Introductory Physics 3.00m...

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