Problem 8 · 2010 AMC 8
Medium
Ratios, Rates & Proportions
relative-speed
As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/2 mile in front of her. After she passes him, she can see him in her rear mirror until he is 1/2 mile behind her. Emily rides at a constant rate of 12 miles per hour, and Emerson skates at a constant rate of 8 miles per hour. For how many minutes can Emily see Emerson?
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Answer: D — 15 minutes.
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Hint 1 of 2
Don't track two moving people — track the gap between them. It starts at 1/2 mile (Emerson ahead) and ends at 1/2 mile (Emerson behind), a swing of 1 full mile.
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Hint 2 of 2
When two things move the same direction, sit in one rider's seat: from Emily's view Emerson drifts backward at the difference of the speeds, 12 − 8 = 4 mph. That's relative speed.
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Approach: watch the gap, using relative speed
- From Emily's seat, Emerson slides backward at 12 − 8 = 4 mph — only the difference matters since they head the same way.
- The gap she can ‘see across’ runs from 1/2 mile ahead to 1/2 mile behind, so 1 mile of relative drift.
- Time = distance / speed = 1 / 4 hour = 15 minutes.
- Why this transfers: same direction → subtract speeds; opposite directions → add them. Reframing two movers as one gap turns chase/overtake problems into a single distance ÷ rate step.
Another way — track real positions and solve:
- Let t be hours since Emily is alongside Emerson. Emily has gone 12t, Emerson 8t, so Emily leads by 4t miles.
- Visibility starts at 4t = −1/2 (he's ahead) and ends at 4t = +1/2 (he's behind), a span of Δ(4t) = 1, i.e. t increases by 1/4 hour.
- 1/4 hour = 15 minutes — same answer, confirming the gap view.
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