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Car a runs at a constant speed on a straight road at a speed of 10m / s. car B moves in a straight line at a speed of 4m / s in the same direction as car A. car a starts to brake at an acceleration of 0.5m/s when passing by car B. the timing starts from car a's braking Before car B overtakes car a, the distance between the two cars is the maximum Why is the maximum distance between two cars when the speed of a and B is equal?
The speed of car a is higher than that of car B. as long as the speed of car a is higher than that of car B, the distance between Party A and Party B will continue to increase. Only when the speed of car a is lower than that of car B, the distance between Party A and Party B will continue to decrease. Therefore, when the speed of car a and Party B is the same, the distance between Party A and Party B is the largest, which is the turning point of the distance reduction between Party A and Party B
When the car is driving on the straight road at the speed of V1 = 10m / s, it suddenly finds that there is a bicycle moving in the same direction at the speed of V2 = 4m / s. if the car immediately turns off the accelerator and decelerates at the acceleration of a = - 6m / S2, the car just does not collide with the bicycle, then how big is s?
The condition that the two do not collide is: when the speed is equal, the displacement of the car is equal to the displacement of the bicycle plus the initial distance. Let the time of the movement be t. from the solution of equal speed: V2 = V1 + at4 = 10-6 × T, t = 1s, the displacement relation is: S + v2t = v1t + 12at12, and the solution is: S = 3M. A: the car just does not collide with the bicycle, s should be 3M
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