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A student runs at 5 m s^-1 to catch a bus starting from rest with 1 m s^-2 acceleration. At what time does the student catch the bus for the first time?
Solving quadratic equation 0.5t^2 – 5t + c = 0 yields first meeting time roots. First meeting occurs at t = 2 or 4 depending on setup.
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A police car starts moving with 5 m s^-2 acceleration 6 s after a car passes by at 20 m s^-1. How many meters ahead does the police car catch the car?
1. x_0 = 120 m. 2. 2.5t^2 – 20t – 120 = 0 -> t = 12 s. 3. Distance = 0.5 * 5 * 12^2 = 360 m.
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Two cars moving at same speed start at t = 0; one accelerates with a, other decelerates with -a. When decelerating car stops with displacement x, what is distance between cars?
1. Decelerating: x = v^2 / (2a). 2. Accelerating: x_acc = 3x. 3. Distance difference = 3x – x = 2x.
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Runner 1 moves at 8 m s^-1 and is 40 m from finish. Runner 2 starts 50 m away with 6 m s^-1 and constant acceleration a. Both reach finish simultaneously. What is a?
1. t = 40 / 8 = 5 s. 2. 50 = 6 * 5 + 0.5 * a * 25 -> 20 = 12.5a -> a = 1.6 m s^-2.