The function relationship between the braking distance s (m) and the speed U (km / h) is s = 1 / 100u & # 178; when a vehicle with a speed of 100 km / h is 80 m in front of it, it is found that the braking distance s (m) is a function of the speed U (km / h)?

The function relationship between the braking distance s (m) and the speed U (km / h) is s = 1 / 100u & # 178; when a vehicle with a speed of 100 km / h is 80 m in front of it, it is found that the braking distance s (m) is a function of the speed U (km / h)?

1, according to the formula: S = 0.01 V ^ 2,
When the speed is 100 km / h, the braking distance is 0.01 × 100 ^ 2 = 100 m
Therefore, it is difficult to avoid finding the target at 80 meters

The function relationship between the braking distance s (m) and the speed U (km / h) is s = 1 / 100u & sup2; when a faulty vehicle is found 80 m in front of a vehicle with a speed of 100 km / h, is it dangerous to brake at this time?

From the meaning of the title, s = (1 / 20) x + (1 / 180) x ^ 2
39.5=(1/20)x+(1/180)x^2
39.5=0.05x+(1/180)x^2
(1/180)x^2-0.05x+39.5=0
Let's solve the quadratic equation of one variable by ourselves

On a certain distance, the distance D is a positive proportional function of the product of the square of the vehicle speed V (km / h) and the body length s, and the minimum distance should not be less than half of the body length. Assuming that the vehicle speed is 50km / h and the distance is exactly equal to the body length, the functional relationship of D with respect to V (s is a constant) is written

The distance D is a positive proportional function of the product of the square of the vehicle speed V (km / h) and the body length s (m). It can be seen that d = kV ^ 2S (k is the proportional coefficient). By assuming that the vehicle distance is exactly equal to the body length when the vehicle speed is 50 km / h, it can be seen that s = k * 50 ^ 2SK = 1 / 2500d = 1 / 2500V ^ 2S, because the minimum distance should not be less than half of the body length

One liter of gasoline can drive a car for 15 kilometers. If the car travels an average of 240 kilometers a day, how much can 320 liters of gasoline drive the car

320 times 15 divided by 240 equals 20 (days)