ch02-p040 - 2 where t 2 = x 2 ⁄ v 2 We simultaneously...

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40. Let d be the 220 m distance between the cars at t = 0, and v 1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x 1 = 44.5 m) and v 2 be the 40 km/h =100/9 m/s speed (corresponding to passing point of x 2 = 76.6 m) of the red car. We have two equations (based on Eq. 2-17): d x 1 = v o t 1 + 1 2 at 1 2 where t 1 = x 1 v 1 d x 2 = v o t 2 + 1 2 at 2
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Unformatted text preview: 2 where t 2 = x 2 ⁄ v 2 We simultaneously solve these equations and obtain the following results: (a) v o = − 13.9 m/s. or roughly − 50 km/h (the negative sign means that it’s along the – x direction). (b) a = − 2.0 m/s 2 (the negative sign means that it’s along the – x direction)....
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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