HW02Sol - Jim Guinns PHYS2211 Assignment #2 Solutions 1....

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Jim Guinn’s PHYS2211 Assignment #2 Solutions 1. Starting at the initial position of +25m , Azadeh walks with a constant velocity of -3m/s . With x o = 25m and v o = -3m/s we have that x = x o + v o t = 25m – (3m/s)t , so a) What is Azadeh’s position after 4sec? x(4s) = 25m – (3m/s) (4s) = 25m – 12m = x = 13m . b) What is Azadeh’s position after 12sec? x(12s) = 25m – (3m/s) (12s) = 25m – 36m = x = -11m . c) What is Azadeh’s velocity after 20sec? x(20s) = 25m – (3m/s) (20s) = 25m – 60m = x = -35m . d) What is Azadeh’s acceleration after 25sec? x(25s) = 25m – (3m/s) (25s) = 25m – 75m = x = -50m . 2. Seeing Azadeh out walking, starting from the origin Wakjira decides to go for a run with a velocity shown in the graph below. What is Wakjira’s position at t = 70.0s? Wakjira is running with a constant velocity, so we can use x = x o + v o t . With x o = 0 (since he starts from the origin) we have x = (20m/s)t = (20m/s) (70s) = x = 1400m . 3. Seeing Azadeh and Wakjira outside, starting from the origin Anna decides to go for a run
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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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HW02Sol - Jim Guinns PHYS2211 Assignment #2 Solutions 1....

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