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The car moves in a straight line at a speed of 20 m / s, and the acceleration after braking is 5 m / S2. Then the ratio of the displacement within 2 s after braking to that within 5 s after braking is () A. 1：1B. 3：1C. 3：4D. 4：3
Then X1 = v0t + 12at21 = (20 × 2-12 × 5 × 22) M = 30m. 5S > 4S, so the displacement in the first 5S is equal to that in the first 4S. Then x2 = v0t + 12at22 = (20 × 4-12 × 5 × 42) M = 40m
When the car runs at a speed of 16m / s and suddenly brakes at an acceleration of 8m / S2, what is the displacement of the car in 3S after braking?
The velocity is 16m / s, and the acceleration is - 8m / S ^ 2. The 2S velocity is 0, so the 3S displacement is the 2S displacement,
S=1/2*8*2^2=16m
What is the speed of the car at the end of the third second after braking if the speed of uniform acceleration of the car is 10m / s and the acceleration of emergency braking is 5m every second in case of red light
Sorry. It should be "the speed of uniform motion."
When the car stops, t = V / a = 2S, that is, the time required for the car to stop braking is 2S
So the car's speed is zero at the end of 3 seconds
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