p02_038

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
38. 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 ` (which is also the Fnal velocity of the train, if the rear-end collision is barely avoided). We note that the distance ∆ x consists of the original gap between them D as well as the forward distance traveled during this time by the locomotive v ` t . Therefore, v t + v ` 2 = x t = D + v ` t t = D t + v ` . We now use Eq. 2-11 to eliminate time from the equation. Thus, v t + v ` 2 = D ( v ` v t ) /a + v ` leads to a = µ v t + v ` 2 v ` ¶µ v ` v t D = 1 2 D ( v ` v t ) 2 . Hence, a = 1 2(0 . 676 km) µ 29 km h 161 km h 2 = 12888 km / h 2 which we convert as follows:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a = ³ − 12888 km / h 2 ´ µ 1000 m 1 km ¶µ 1 h 3600 s ¶ 2 = − . 994 m / s 2 so that its magnitude is 0 . 994 m/s 2 . A graph is shown below for the case where a collision is just avoided ( x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight) line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. The other case (where the colli-sion is not quite avoided) would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet. 200 400 600 800 x 10 20 30 t...
View Full Document

Ask a homework question - tutors are online