(a) What is the maximum distance the train can travel if it accelerates from res
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(a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes? got 22.913 (b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions? CAN'T UNDERSTAND HOW TO GET (c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart. Is 30.55? (d) The trip from one station to the next takes at minimum 37.5 minutes. How far apart are the stations? 55.425?? Can you explain in simple terms how to do b and d?Explanation / Answer
First of all, the important thing to understand is that acceleration equals the change in velocity over the change in time. Or, in calculus terms: a = dv / dt Velocity equals the change in distance over the change in time. V = dx / dt From that, we know that if we integrate "a" with respect to time, we'll get "v". And if we integrate "v" with respect to time, we'll get "x". The integral of "a" with respect to time in this case is easy because "a" is a constant of 3 ft/s2. The integral of "a" (which is velocity) is "3t + vi" where "vi" equals the initial velocity. Now we integrate this equation to get the final equation. The integral of that equation is "1.5*t^2 + vi*t + xi" where xi is the initial distance traveled. This equation describes the distance an object travels while it is accelerating: x = 1.5*t^2 + vi*t + xi A. The object accelerates until it reaches a final velocity of 110 mi/hr, which equals 161.3 ft/s. a = dv / dt 3 = (161.3 - 0) / dt dt = 53.8 s Therefore, it takes 53.8 s to accelerate to 110 mi/hr. Now we can use the equation above to determine how far it travelled during those 53.8 s. x = 1.5*t^2 + vi*t + xi vi = 0 and xi = 0 x = 1.5*t^2 = 1.5 * 53.8^2 = 4338 ft = .822 miles The problem then says that the train runs at 110 mi/hr for 15 minutes (0.25 hours). Therefore: v = dx / dt 110 = dx / .25 dx = 27.5 miles Total distance = 0.822 + 27.5 = 28.3 miles B. We know that the train takes 53.8 s and 0.822 miles to accelerate to max velocity. We also know the train deccelerates at the same speed and, henceforth, uses the same distance. The train will take a total of 107.6 seconds (2*53.8 seconds) to stop and slow down. The rest of the 15 minutes (900 seconds) will be spent travelling at 110 mi/hr. v = dx / dt 161.3 = dx / (900 - 2*53.8) dx = 127848 feet = 24.2 miles Total distance traveled = 24.2 + 2*.822 = 25.9 miles C. We know that it takes a total of 1.64 miles (2*.822 miles) to accelerate and deccelerate. Therefore, the rest of the 55 miles will be traveled at constant speed. v = dx / dt 110 = (55 - 2*.822) / dt dt = 0.49 hrs = 29.1 minutes Total time = 29.1 + 2*53.8/60 = 30.9 minutes D. We know that it takes a total of 107.6 seconds to accelerate and deccelerate (2*53.8 seconds). The rest of the travelling time will be spent travelling at 110 miles per hour. v = dx / dt 161.3 = dx / (37.5*60 - 107.6) dx = 345,648 ft = 65.46 miles Total distance = 65.46 + 2*.822 = 67.1 miles
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