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A motorcycle and a car start from rest from the same place at the same and travel in the same direction. The motorcycle acceleration at 1.0 m s^(-1) up to a speed of 36 km h^(-1) and the car at 0.5 m s^(-1) up to a speed of 54 km h^(-1). The time at which the car would overtake the motorcycle is |
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Answer» 20 s For motor cycle : Maximum speed attained = `36 km H^(-1) = 36 xx 5/18 = 10 m s^(-1)` Since its acceleration = `1.0 m s^(-2)`, the time `t_1` taken by it to attain the maximum speed is given by `v = u + at_1 implies 10 = 0 + 1.0 xx t_1 implies t_1 = 10 s`(sinceu = 0) The distance covered by motocycle in attaining the maximum speed is `S_1 = 0 + 1/2 at^(2)_1 = 1/2 xx 1.0 xx (10)^(2)` = 50 m The time during which the motorcycle moves with maximum speed is (t - 10) s. The distance covered by the motorcycle during the time is `S^(')_1 = 10 xx (t - 10) = (10t - 100)` m `therefore` Total distance travelled by motorcycle in time t is `S = S_1 + S^(')_1 = 50 + (10t - 100) = (10t - 50) m`...(i) For car : Maximum speed attained = `54 km h^(-1) = 54 xx 5/18 = 15 m s^(-1)` Since its acceleration = `0.5 m s^(-2)` The time taken by it to attain the maximum speed is given by 15 = `0 + 0.5 xx t_2` or `t_2 = 30 s`(since u = 0) The distance covered by the car in attaining the maximum speed is `S_2 = 0 + 1/2 at^(2)_2 = 1/2 xx 0.5 xx (30)^(2)` = 225 m The time during which the car moves with maximum speed is (t - 30) s. The distance covered by the car during this time is `S^(')_2 = 15 xx (t - 30) = (15t - 450)` m `therefore` Totoal distance travelled by car in time t is `S = S_2 + S^(')_2 = 225 + (15t - 450) = (15t - 225)`m...(II) From equations (i) and (ii), we GET `10t - 50 = 15t - 225` or `5t = 75` or t = 35 s |
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