Help me please?
m/2*V^2 = F*d d = m*V^2/(2*F) = 2.10*10^5*12^2/(7.4*10^4) = 408.6 m d = V*t/2 t = d/(V/2) = 408.6/6 = 68.108 sec
a) How long does it take the train to stop? Let’s use the following equation to determine the train’s acceleration. a = Force ÷ Mass = -3.70 * 10^4 ÷ 2.10 * 10^5 = -3.7 ÷ 21 The acceleration is approximately -0.176 m/s^2. Let’s use the following equation to determine the time for the train’s velocity to decrease from 12 m/s to 0 m/s. vf = vi + a * t 0 = 12 + (-3.7 ÷ 21) * t t = 12 ÷ (3.7 ÷ 21) = 252 ÷ 3.7 The time is approximately 68 seconds. b) How far does the train travel during this time? Let’s use the following equation to determine the distance the train moves. d = ½ * (vi + vf) * t, vf = 0 d = ½ * 12 * 252 ÷ 3.7 = 1,512 ÷ 3.7 The distance is approximately 410 meters. These two answers are rounded to two significant digits. I hope this helps you to understand how to solve this type of problem.
Dear Sydney, You have a solid answer from a tip top Y! Answer man. I am moving your question over to where rail road operators are lurking about, thirsty for this very question. They look at and solve this problem every time they pull out on a freight train. Not so much with passenger service, because those trains are built with standard cars, engines and lengths. With the exceptions of coal, rock, or oil trains, freight trains in general are quite complicated in their weight distribution. Let's see what those men say about this one, OK?