Question:
Ship is 18 miles from shore when a seaplane, 10x faster, is sent. Where does it catch up?
Ship is 18 miles from shore when a seaplane, 10x faster, is sent. Where does it catch up?
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Use relative speed and equate distances for both moving bodies.
Updated On: Aug 6, 2025
- 24 miles
- 25 miles
- 22 miles
- 20 miles
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The Correct Option is D
Solution and Explanation
Let ship’s speed = \( v \), plane’s speed = \( 10v \)
Let \( t \) be the time after which they meet. Ship travels: \( 18 + vt \)
Plane travels: \( 10v \cdot t \) Set equal: \[ 18 + vt = 10vt 18 = 9vt t = \frac{2}{v} \] Distance from shore: \[ v \cdot \frac{2}{v} = 2 18 + 2 = \boxed{20 \text{ miles}} \]
Let \( t \) be the time after which they meet. Ship travels: \( 18 + vt \)
Plane travels: \( 10v \cdot t \) Set equal: \[ 18 + vt = 10vt 18 = 9vt t = \frac{2}{v} \] Distance from shore: \[ v \cdot \frac{2}{v} = 2 18 + 2 = \boxed{20 \text{ miles}} \]
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