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Zorluk: KolayAlgebraic Word Problems and Modeling

Train XX departs from a station traveling due east at a constant speed of 5050 miles per hour. Exactly 11 hour later, Express Train YY departs from the same station along the same track, traveling due east at a constant speed of 7575 miles per hour. How many hours after Express Train YY departs will it catch up to Train XX?

  1. A
    11
  2. B
    1.51.5
  3. 22Cevap
  4. D
    2.52.5
  5. E
    33

Cevap

Express Train YY will catch up to Train XX exactly 22 hours after Express Train YY departs.
Let tt represent the number of hours Express Train YY travels. Because Train XX departed 11 hour earlier, it has been traveling for t+1t + 1 hours. For Express Train YY to catch up to Train XX, both trains must cover the exact same distance from the starting station. Setting up the distance equation 75t=50(t+1)75t = 50(t + 1) gives 75t=50t+5075t = 50t + 50, which simplifies to 25t=5025t = 50, yielding t=2t = 2 hours.

Adım Adım Çözüm

1
Define variables for the time traveled by each train.
Let tt be the time in hours that Express Train YY travels. Since Train XX departed 11 hour earlier, Train XX travels for t+1t + 1 hours.
Train XX has a 11-hour head start.
2
Express the distance traveled by each train using Distance=Rate×Time\text{Distance} = \text{Rate} \times \text{Time}.
Distance of Train X=50(t+1)X = 50(t + 1) miles; Distance of Express Train Y=75tY = 75t miles.
Both trains travel at constant rates along the same path.
3
Equate the two distance expressions to solve for tt.
75t=50(t+1)    75t=50t+50    25t=50    t=275t = 50(t + 1) \implies 75t = 50t + 50 \implies 25t = 50 \implies t = 2.
Express Train YY catches Train XX when both have covered the exact same distance.

Anahtar Kavram

Distance, Rate, and Time Modeling for Catch-up Scenarios
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