HW1-phy2048-Spring-2014-solution

25 hour and the distance that remains is 160 km thus

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/24/2014.

Ask a homework question - tutors are online