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A bus starts from rest with acceleration 1 m/s². At the same instant, a person standing 11 m behind the bus begins running at a constant speed of 5 m/s to catch the bus. How long will it take the person to catch the bus?
- 5 + sqrt(3) sec
- 5 - 2*sqrt(3) sec
- 15 - sqrt(3) sec
- 5 - sqrt(3) sec
Correct answer: 5 + sqrt(3) sec
Solution
Position of person (starting from 0): xₚ = 5t. Position of bus (starting from 11 m): x_b = 11 + (1/2)*t². Setting equal: 5t = 11 + t²/2 => t²/2 - 5t + 11 = 0 => t² - 10t + 22 = 0 => t = [10 +/- sqrt(100-88)]/2 = [10 +/- sqrt(12)]/2 = 5 +/- sqrt(3). Both solutions are valid catching times. The first catch at t = 5 - sqrt(3) approx 3.27 sec, second at t = 5 + sqrt(3) approx 6.73 sec. The question asks how long it takes to catch (first time), so t = 5 - sqrt(3) sec.
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