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When a bus comes into the station, it starts to brake and make a uniform deceleration linear motion. The displacement in the first second and the second second is 9m and 7m respectively, then the displacement in the sixth second after braking is () A. 20 mB. 24 mC. 25 mD. 75 m
Let V 0 be the initial velocity and a be the acceleration of the car; X2-x1 = at2, a = x2 − x1t2 = 7 − 912 = - 2m / S2. According to the displacement in 1s: X1 = v0t + 12at2, substituting into the data, 9 = V0 × 1 + 12 × (− 2) × 12, the solution is V0 = 10m / s. The time to stop is t0 = 0 − v0a = 0 − 10 − 2S = 5S. Then the displacement in 6S after braking is equal to the displacement in 5S, which is x = v02t0 = 102 × 5m = 25m
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