# ch02-p038 - being a couple of meters into the intersection...

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() 2 22 0 2 (15.28 m/s) 2 25 . 1 8 m / s bb v v ad d =+ ¡ =− which yields d b = 22.53 m. Thus, the total distance is d r + d b = 34.0 m, which means that the driver is able to stop in time. And if the driver were to continue at v 0 , the car would enter the intersection in t = (40 m)/(15.28 m/s) = 2.6 s which is (barely) enough time to enter the intersection before the light turns, which many people would consider an acceptable situation.
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Unformatted text preview: being a couple of meters into the intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too long (the light turns in 1.8 s). The driver is caught between a rock and a hard place. 38. We take the direction of motion as + x , so a = –5.18 m/s 2 , and we use SI units, so v = 55(1000/3600) = 15.28 m/s. (a) The velocity is constant during the reaction time T , so the distance traveled during it is d r = v T – (15.28 m/s) (0.75 s) = 11.46 m. We use Eq. 2-16 (with v = 0) to find the distance d b traveled during braking:...
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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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