Question

6. A 3000 lbm car is moving initially at a speed of 20 ft/s south bound towards an intersection with a railroad track. When the car is 100 ft from the intersection, the driver sees a train traveling east bound toward the intersection at a speed of 15 ft/s. At the same same instant the train is 40 ft from the intersection. The train maintains a constant speed, and the driver has negligible reaction time. If the car is capable of accelerating at a rate of a(t) = m/s?, can the car beat the train to the interesection? Support your answer mathematically. Car 20 ft/s 100 ft Train 40 ft 15 ft/s

Answer 1

The detailed step by step calculations for the given problem is described in the images below:

Page 1 Coor 2oft/s 100ft Traint 40 ft 15ft/s Train is travelling at a constant speed, Uq = 15 ft/s so, cit this constant speed, time required for train to reach an intersection which is 40 ft fus from the intersection is given by, "T distance (s) time ct) it 40 15 t = 2.67 Sec. Now, car is moving initially cat speed, uc = zoft/s Distance between the car and an intersection at this instance of time, d = 100 ft. Now, if a cu wishes to reach an intersection which is d=looft fas from the car in the time t=2.67 see, the acceleration of the cas should be such that, d =ų.t + Ź qet²

Perge-2 .:100 = (20).(2.67)+ ļ4.(2.67)2 - 13.0735 ft/s² a - 3.9848 m/s² So, now If the car wants to beat the train at the intersection, the acceleration of the cas must be greater than q (3.9848 m/s²). But the car is capable of accelerating at a rate of, acto lo t m/s² For t=2.67 sec, act) To 3.7453 m/? 2.67 a : The car is accelerating at rate of act) = 3.7453 m/s² at time t=2.67 sec. which is less than the required accelerating rate of the cas to beat the train at the enter intersection cach. ie. act) < (ach required The car cannot beat the train to the intersection. Ans E