Help please! A train leaves the station........?

Question

A train leaves the station traveling due east at a rate of 45 miles per hour.
A second train leavesa diffferent station an hour later traveling due west on the same track, going 60 miles per hour

If the stations are 255 miles apart, how many miles from the halfway point between the stations will the trains collide?

A) 0 B) 5 C) 7.5 D)12.5 E) 25

PLease help! I dont know where to start.... thank you!!!

Answer 1

The two trains are traveling towards each other at a combined speed of 105 mph. Since the first train traveled 45 miles in the first hour (before the 2nd train left the station), the distance between the trains after the first hour is 255-45 = 210 miles.

At 105 combined mph, the two trains will collide in two hours (210/105 = 2). In those two hours the second train will have traveled 120 miles from its station (2 x 60). Since the midpoint between the two stations is 255/2 = 127.5, the trains will be 7.5 miles from the midpoint when they collide. The answer is C) 7.5.

Best wishes and good luck.

Answer 2

the respond is a, 7:fifty six am Assuming that the full distance is a million kilometer or one thousand meter, the linked fee of practice A is 1000m/4 hours = 1000m/240 minutes or 4.1667 meters/minute the linked fee of practice B is 1000m/3.5 hours = 1000m/210 minutes or 4.7619 meters/minute At 7:00 am, all of us understand that practice A is already halfway to Station B, mutually as practice B is stiil table sure. Distance = velocity multiply with the help of time ( D = S x T) At 7:40 5 am, practice A would be at distance 4.1667m/minute x one hundred sixty five minutes ( 2 hours & 40 5 minutes) = 687.50 meters from Station A, mutually as practice B would be at 4.7619 m/minutes x 40 5 minutes = 214.285m from Station B or 785.715 meters ( one thousand meter - 214.285 meter) from Station A, meaning they did not meet at 7:40 5 am. At 7:fifty six am., practice A would be at 4.1667 meters/minute x 176 minutes ( 2 hours & fifty six minutes) = 733.33 meters from Station A. At 7:fifty six am, practice B would be at 4.7619 meters/minute x fifty six minutes = 266.sixty seven from Station B or 733.33 meters (one thousand meters - 266.sixty seven meters) from Station A. that's today - 7:fifty six am, that the two trains met one yet another. today, i'm nevertheless determining the sumptuous equation, if there is.