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 vtand the locomotive’s velocity as vA(which is also the final velocity of the train, if the rear-end collision is barely avoided). We note that the distanceΔxconsists of the original gap between them Das well as the forward distance traveled during this time by the locomotivev tA. Therefore, vvxtDv ttDtvt+==+=+AAA2Δ.We now use Eq. 2-11 to eliminate time from the equation. Thus, vvDvvavtt+=−+AAA2bg/which leads to avvvvvDDvvttt=+−FHGIKJ−FHGIKJ= −−AAAA2122bg.Hence, a= −−FHGIKJ= −12 0 676291611288822(.km)kmhkmhkm / h which we convert as follows:
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