Keep in mind I just started Grade 11 Physics a monthago. A freight train is roll
ID: 1670216 • Letter: K
Question
Keep in mind I just started Grade 11 Physics a monthago. A freight train is rolling down the track at 65.0km/hr. Out ofthe fog, 1km behind, a fast express going 120km/hr, appears on thesame track. The Express engineer slams on the brakes. With thebrakes set, he needs 3.00km to stop. Will the express rear-end thefreight train or will the express train be able to AVOID acrash? I'm not really sure about how to go about this problem. Ineed this for tommorow because it's worth 5 percent of my totalmark, so it's a tough situation. This isn't just some homeworkproblem. Keep in mind I just started Grade 11 Physics a monthago. A freight train is rolling down the track at 65.0km/hr. Out ofthe fog, 1km behind, a fast express going 120km/hr, appears on thesame track. The Express engineer slams on the brakes. With thebrakes set, he needs 3.00km to stop. Will the express rear-end thefreight train or will the express train be able to AVOID acrash? I'm not really sure about how to go about this problem. Ineed this for tommorow because it's worth 5 percent of my totalmark, so it's a tough situation. This isn't just some homeworkproblem.Explanation / Answer
first we need to know the time it takes for the express train tobrake. we know vf = 0 km/h vi = 120km/h xf = 3 km xi = 0 km vf = vi + at 0 = 120 + at at = -120 t = -120/a xf = xi + vi(t) + 1/2at^2 3 = 0 + 120(-120/a) + 1/2a(-120/a)^2 3 = -14400/a - 7200a/a^2 3 = -14400/a - 7200/a 3 = -21600/a a = -7200 km/h^2 so t equals t = -120/a = -120/-7200 = 0.01666 h] once we know the time, the distance the freight train has moved bythe time the express train completely stops is xf = xi + vt + 1/2at^2 = 1 + 65(0.01666) + 0 = 1+ 1.08333 =2.083333 km since this number is less than the 3 km nessacery for the expresstrain to stop, the express train will rear-end the freighttrain. Hopefully this helps!
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