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HW2 solutions

# HW2 solutions - HomeWork 2 solutions 1 Cutnell Johnson 8th...

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HomeWork 2 solutions Physics 203 1. Cutnell & Johnson 8 th Edition, Chapter 2, Problem 67 REASONING The slope of a straight-line segment in a position-versus-time graph is the average velocity. The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope. SOLUTION a. The slope, and hence the average velocity, is positive for segments A and C , negative for segment B , and zero for segment D . b. In the given position-versus-time graph, we find the slopes of the four straight-line segments to be 1.25 km 0 km 6.3 km/h 0.20 h 0 h 0.50 km 1.25 km 3.8 km/h 0.40 h 0.20 h 0.75 km 0.50 km 0.63 km/h 0.80 h 0.40 h 0.75 km 0.75 km 0 km/h 1.00 h 0.80 h A B C D v v v v = = + = = = = + = = 2. Problem 39. Add “during this time” to the end of the last sentence. REASONING Because the car is traveling in the + x direction and decelerating, its acceleration is negative: a = −2.70 m/s 2 . The final velocity for the interval is given ( v = +4.50 m/s), as well as the elapsed time ( t = 3.00 s). Both the car’s displacement x and its initial velocity v 0 at the instant braking begins are unknown. Compare the list of known kinematic quantities ( v , a , t ) to the equations of kinematics for constant acceleration: 0 v v at = + (Equation 2.4), ( ) 1 0 2 x v v t = (Equation 2.7), 2 1 0 2 x v t at = + (Equation 2.8), and 2 2 0 2 v v ax = + (Equation 2.9). None of these four equations contains all three known quantities and the desired displacement x , and each of them contains the initial velocity v 0 . Since the initial velocity is neither known nor requested, we can combine two kinematic equations to eliminate it, leaving an equation in which x is the only unknown quantity.

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