(1 / 2) the car starts to move at an acceleration of 1 meter every second. At 20 meters behind the car, while the car starts to move, someone starts to ride a bicycle at the same time (1 / 2) the car starts to move forward at an acceleration of 1 meter per second. 20 meters behind the car, while the car starts to move, someone starts to chase the car at a constant speed of 6 meters per second. Can he catch up with the car (1 / 2) the car is driving at the speed of 18 meters per second. When the driver suddenly finds an obstacle in front of him, he brakes immediately. Assuming that the acceleration is 3 meters per second after braking, he decelerates evenly, and calculates the sliding distance of the car in 10 seconds (2 / 2)?

Let t be the time to catch up
6*t-1/2at^2=20
T has no solution, so it can't catch up
When the speed is equal, the distance between two cars is the minimum
6=at
t=6s
s1=6*6=36m
s2=1/2at^2=18
Minimum distance = 20 + 18-36 = 2m

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