Asked by Kristina
A train leaves the station at 8 a.m. and averages 40 miles per hour another train leaves the same station one hour later and averages 50 miles per hour traveling in the same direction on a parallel track at what time will the second train catch up with the first train how many miles would each train have traveled by that time
Answers
Answered by
Reiny
Time taken by the slower train ---- t hrs
time taken by faster train --------- t - 1 hrs
distance gone by slower train = 40t miles
distance gone by faster train = 50(t-1) miles
But, when they pass each other, each will have covered the same distance, so
50(t-1) = 40t
50t - 50 = 40t
10t = 50
t = 5
Since the slower train left at 8:00 am and it took 5 hours for them to meet, they met at 1:00 pm
The distance they both went is 200 miles
time taken by faster train --------- t - 1 hrs
distance gone by slower train = 40t miles
distance gone by faster train = 50(t-1) miles
But, when they pass each other, each will have covered the same distance, so
50(t-1) = 40t
50t - 50 = 40t
10t = 50
t = 5
Since the slower train left at 8:00 am and it took 5 hours for them to meet, they met at 1:00 pm
The distance they both went is 200 miles
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