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HW1-phy2048-Spring-2014-solution

# 25 hour and the distance that remains is 160 km thus

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Unformatted text preview: Ddown 2D = D D t up + tdown + vup vdown 2 PHY2048 Spring 2014 HW1 which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the average speed. 2P6. Huber’s speed is v0 = (200 m)/(6.509 s) =30.72 m/s = 110.6 km/h, where we have used the conversion factor 1 m/s = 3.6 km/h. Since Whittingham beat Huber by 19.0 km/h, his speed is v1 = (110.6 km/h + 19.0 km/h) = 129.6 km/h, or 36 m/s (1 km/h = 0.2778 m/s). Thus, using Eq. 2- 2, the time through a distance of 200 m for Whittingham is Δ x 200 m Δt = = = 5.554 s. v1 36 m/s 2P11. The book values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in...
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