An express train, traveling at 36.0 m/s, is accidentally sidetracked onto a loca

Question

An express train, traveling at 36.0 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a local train exactly 1.00 x 102 m ahead on the same track and traveling in the same direction. The local engineer is unaware of the situation. The express engineer jams on the brakes and slows the express at a constant rate of 3.00 m/s2. If the speed of the local train is 11.0 m/s, will the express train be able to stop in time or will there be a collision? To solve this problem, take the position of the express train when it first sights the local train as a point of origin. Next, keeping in mind that the local train has exactly a 1.00

Explanation / Answer

dexpress = v0t + 1/2 at2

           = (36.0 m/s)(12.0 s)+1/ 2. (

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