A 5-meter ladder leans against a vertical wall. If the bottom of the ladder slips away from the wall at a rate of 1.5 m/s, how fast is the top of the ladder sliding down the wall when the bottom is 3 meters from the wall?
Answer and explanation
Correct answer: C. 1.125 m/s (or 9/8 m/s)
- Step 1: Let x be the distance from the wall to the base, and y be the height on the wall. By the Pythagorean theorem: x^2 + y^2 = 5^2 = 25.
- Step 2: When x = 3, y = sqrt(25 - 9) = sqrt(16) = 4 meters.
- Step 3: Differentiate with respect to time t: 2x(dx/dt) + 2y(dy/dt) = 0 => x(dx/dt) + y(dy/dt) = 0.
- Step 4: Substitute values: 3(1.5) + 4(dy/dt) = 0 => 4.5 + 4(dy/dt) = 0 => dy/dt = -4.5 / 4 = -1.125 m/s. The sliding speed is 1.125 m/s.
Answer
1.125 m/s (or 9/8 m/s).
Why not the others: '2.0 m/s' assumes dy/dt = -dx/dt * (4/3); '1.5 m/s' assumes the top slides at the same speed as the bottom; '0.75 m/s' divides by 6 instead of 4.
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