Solution - American Association of Physics Teachers

2013 F = ma Exam 21. An observer stands on the side of the front of a stationary train. When the train starts moving with constantacceleration, it takes 5 seconds for the first car to pass the observer. How long will it take for the 10th car to pass?(A) 1.07s(B) 0.98s(C) 0.91s(D) 0.86s(E) 0.81sSolutionStart with∆x = 1 2 at2 + v i tWe have four times. t 0 = 0 is when the train starts, and when the first car is aligned with the observer. t 1 iswhen the end of the first car is aligned with the observer. ThenL = 1 2 at 1 2We are assuming the car has a length L. t 2 is when the tenth car is first aligned with the observer, so9L = 1 2 at 2 2and finally, t 3 is when that car has passed,10L = 1 2 at 3 2From the equation for t 1 we findsoand2L/a = 25 s 2t 2 = √ 9 · 25 s 2 = 15 st 3 = √ 10 · 25 s 2 = 15.81 sCopyright c○2013 American Association of Physics Teachers