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PHYS 1301 - GENERAL PHYSICS I Tuesday/Thursday 10:50AM-12:05PM - Dr. W. Anderson Solutions to Homework from Chapter 2 Problem 7 Problem: A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (b) the average velocity, in m/s. Solution: The total distance that the jogger covers is 8 lap · 0.25 mile/lap = 2 miles. However, the joggers total displacement is 0 miles, since she finishes at the same place she starts. (a) Since speed is distance/time, the jogger’s speed is 1610 m 1 min 2.0 mi · · = 4.3 m/s. 12.5 min mi 60 s (b) Since the jogger’s total displacement is 0.0 m, her average velocity is also 0.0 m. Problem 16 Problem: A sports car is advertised to be able to stop in a distance of 50 m from a speed of 90 km/h. What is its acceleration in m/s2 ? How many g’s is this (1 g = 9.80 m/s2 )? Solution: (a) We need to relate a displacement (50 m), a final velocity (0 m/s, since the car is stopping) and an initial velocity (90 km/h) to an acceleration. The formula which does that is v 2 = v02 + 2a(x − x0 ). First, let us rearrange this equation to isolate a. First, we subtract v02 from both sides, so that we have v 2 − v02 = v02 + 2a(x − x0 ) − v02 = 2a(x − x0 ). Next, we want to divide both side by 2(x − x0 ), 2a(x − x0 ) v 2 − v02 = = a. 2(x − x0 ) 2(x − x0 ) Next, we need to express the initial velocity in m/s, v0 = 90 km/h · 1h · 1000 m/km = 25 m/s. 3600 s Finally, we can put v, v0 and x − x0 into our equation, a= (0 m/s)2 − (25 m/s)2 = 6.25 m/s2 . 2 · 50m (b) To convert this to g’s, we simply multiply a = 6.25 m/s2 · 1 1g 0.64g. 9.80 m/s