Comments on: How much time is required for the entering car to catch up with the other car? http://www.cargearusa.com/blog/how-much-time-is-required-for-the-entering-car-to-catch-up-with-the-other-car/ Fri, 18 May 2012 02:33:04 +0000 http://wordpress.org/?v=2.7 hourly 1 By: Jim http://www.cargearusa.com/blog/how-much-time-is-required-for-the-entering-car-to-catch-up-with-the-other-car/comment-page-1/#comment-10184 Jim Sun, 18 Jul 2010 12:04:58 +0000 http://www.cargearusa.com/blog/how-much-time-is-required-for-the-entering-car-to-catch-up-with-the-other-car/#comment-10184 t = time to catch up constant speed car distance traveled in (t) = 68.9t velocity of accelerating car at main speedway entrance = (3.8)(5.4) = 20.5 m/s accelerating car distance traveled in (t) = Vi(t) + 1/2(5.4)t² = 20.5t + 2.7t² set distances traveled by the two cars equal 68.9t = 20.5t + 2.7t² cancel t's 68.9 = 20.5 + 2.7t 2.7t = 68.9 - 20.5 = 48.4 t = 48.4/2.7 = 17.9 s ANS<a href="http://www.prohomeschool.com/homeschool-driver-ed.htm"> Jim</a> t = time to catch up
constant speed car distance traveled in (t) = 68.9t
velocity of accelerating car at main speedway entrance = (3.8)(5.4) = 20.5 m/s
accelerating car distance traveled in (t) = Vi(t) + 1/2(5.4)t² = 20.5t + 2.7t²

set distances traveled by the two cars equal
68.9t = 20.5t + 2.7t²
cancel t’s
68.9 = 20.5 + 2.7t
2.7t = 68.9 - 20.5 = 48.4
t = 48.4/2.7 = 17.9 s ANS Jim

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