Math Problem Statement

Steven stands on a bridge AB. His distance from A is 7 16 of the distance AB. A train is coming toward A at a speed of 80 km/h. Steven has two choices. He can either run toward A and he will meet the train at A, or he can run toward B at the same speed then the train will catch up him at B. Find his speed in km/h.

Solution

To solve this problem, let the total distance between points AA and BB be dd. Given that Steven is standing 716\frac{7}{16} of the distance from AA, his distance from point AA is 716d\frac{7}{16}d, and his distance from point BB is 916d\frac{9}{16}d.

Let Steven's speed be vv km/h. The speed of the train is given as 80 km/h.

Case 1: Running toward AA

When Steven runs toward AA, the distance he needs to cover is 716d\frac{7}{16}d. Since he and the train meet at AA, the time taken for both will be the same.

The time taken by Steven to reach AA is: t1=716dvt_1 = \frac{\frac{7}{16}d}{v}

The time taken by the train to reach AA is: t2=d80t_2 = \frac{d}{80}

Since t1=t2t_1 = t_2, we have: 716dv=d80\frac{\frac{7}{16}d}{v} = \frac{d}{80}

Canceling out dd from both sides: 716v=180\frac{7}{16v} = \frac{1}{80}

Solving for vv: v=7×8016=35km/hv = \frac{7 \times 80}{16} = 35 \, \text{km/h}

Case 2: Running toward BB

When Steven runs toward BB, the distance he needs to cover is 916d\frac{9}{16}d. The train will also cover this same distance by the time it catches up to him.

The time taken by Steven to reach BB is: t3=916dvt_3 = \frac{\frac{9}{16}d}{v}

The time taken by the train to reach BB is: t4=d80t_4 = \frac{d}{80}

Since t3=t4t_3 = t_4, we have: 916dv=d80\frac{\frac{9}{16}d}{v} = \frac{d}{80}

Canceling out dd from both sides: 916v=180\frac{9}{16v} = \frac{1}{80}

Solving for vv: v=9×8016=45km/hv = \frac{9 \times 80}{16} = 45 \, \text{km/h}

Conclusion

The speed of Steven vv in both cases is 35km/h35 \, \text{km/h} and 45km/h45 \, \text{km/h}. Therefore, the answer is 35 km/h.

Would you like more details or have any questions?

Here are some related problems:

  1. If the train's speed were doubled, what would be Steven's new running speed?
  2. How would the problem change if Steven started at the midpoint of the bridge?
  3. What would happen if Steven could only run at half his current speed?
  4. How long would it take the train to reach Steven if he remained stationary?
  5. If Steven had a head start, how much time would he need to reach safety at point A?

Tip: When solving time-speed problems, always relate the distance covered to the time taken, as time remains constant in uniform motion scenarios.

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Math Problem Analysis

Mathematical Concepts

Speed, Distance, Time Problems

Formulas

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Theorems

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Suitable Grade Level

High School