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A car is driving at a certain speed. The driver finds that there is a danger ahead. It is convenient to stop and brake. After 3S, the car brakes A car is driving at a certain speed. The driver finds that there is a danger ahead. It is convenient to stop and brake. After 3S, the speed of the car before braking is more than kilometers per hour?
Speed at 3 seconds: 10-2 * 3 = 4 m / S
It should be 60 kilometers an hour
The speed of the car on the expressway is 108km / h. If the driver finds a traffic accident 80m ahead, he will brake and make a uniform deceleration linear movement with constant acceleration, and it will take 4S to stop. If the driver's reaction time is 0.5s, try to analyze whether the car will have safety problems
108km/h=30m/s．
Then the displacement in the reaction time X1 = VT1 = 30 × 0.5m=15m．
Displacement of uniform deceleration linear motion x2 = v
2t＝30
two × 4m＝60m．
Because x = X1 + x2 = 75m ＜ 80m
So there will be no safety problem
A: there will be no safety problems with cars
On Shanghai Nanjing Expressway, a car was driving at a constant speed of 108km / h. The driver suddenly found an emergency ahead. After 0.6s, he began to brake, and after 4.4S, he taxied for 52m, and the car stopped. The average speed of the car was ___
∵v=108km/h=30m/s，
The driving distance of 0.6s vehicle is s = VT = 30m / s × 0.6s=18m；
During the process from finding the situation to stopping the vehicle, the vehicle passes through: s' = S + s slip = 18m + 52m = 70m;
The time used is: t ′ = t + T slip = 0.6s + 4.4S = 5S;
The average speed from discovery to stopping is: V ′ = s ′
t′=70m
5s=14m/s；
So the answer is: 14m / s
The car was traveling at a constant speed of 25 meters per second. The driver suddenly found that there was an emergency ahead. After 0.5 seconds (the reaction time of the driver from the beginning to the beginning of braking), the car began to brake automatically, and then taxied for 60 meters after 4.5 seconds. What was the average speed of the car during the period from discovery to complete stop?
60＋25÷2＝145／3≈16.1
The car was traveling at a constant speed of 25 meters per second. The driver suddenly found an emergency ahead. After 0.5 seconds, he began to brake, and after 4.5 seconds, he taxied for 60 meters before the car stopped. The average speed of the car from the time the driver found the situation to the time when it stopped completely The reaction distance is 25 * 0.5 = 12.5m Average speed = (60 + 12.5) / 5 = 14.5 Why should we first calculate the distance through which the reaction passes, and then calculate the average speed, is to add 12.5
Because the question is about the average speed from discovery to complete stop
At this time, the driving distance is 12.5 + 60 = 72.5
The elapsed time is 0.4 + 4.5 = 5
Average velocity = displacement / time
So
The car was traveling at a constant speed of 25 meters per second. The driver suddenly found an emergency ahead. After 0.5 seconds, he began to brake, and after 4.5 seconds, he taxied for 60 meters before the car stopped Ask: the distance the car passes in the driver's reaction time Ask: the average speed of the car from the time the driver finds the situation to the time when it stops completely
Reflecting the passing distance of time s = 0.5 * 25 = 12.5m
Average speed v = 60 / (4.5 + 0.5) = 12m / S
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