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# ch02-p042 - 42 In this solution we elect to wait until the...

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42. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity as v t and the locomotive’s velocity as v A (which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distance Δ x consists of the original gap between them D as well as the forward distance traveled during this time by the locomotive v t A . Therefore, v v x t D v t t D t v t + = = + = + A A A 2 Δ . We now use Eq. 2-11 to eliminate time from the equation. Thus, v v D v v a v t t + = + A A A 2 b g / which leads to a v v v v v D D v v t t t = + F H G I K J F H G I K J = − A A A A 2 1 2 2 b g . Hence, a = − F H G I K J = − 1 2 0 676 29 161 12888 2 2 ( . km) km h km h km / h which we convert as follows:
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