Seven supplementary answers to mathematics.

When driving, it takes some time from finding an emergency to stepping on the brakes, which is called reaction time. After that, it will continue to drive for a certain distance. We call the distance from the discovery of an emergency to parking "braking distance" (pictured).

It is known that there is a relationship between the braking distance S (unit: meter) and the vehicle speed V (unit: meter/second): s=tv+kv2, where T is the driver's reaction time (unit: second) and K is the braking coefficient. In order to test the change of braking distance of drivers after drinking, a certain institution conducted a "drunk" driving test on a certain type of car, and it was known that the braking coefficient of this type of car was k = 0.66.

(1) If volunteers don't drink and the speed is 10 m/s, the braking distance of the car is 15.

Rice;

(2) After drinking a bottle of beer for half an hour, the volunteers drove at the speed of 15m/s, and the braking distance was 52.5m At this time, the response time of the volunteers was 2.

second

(3) If volunteers drive at the speed of 10 m/s, how much will the braking distance increase compared with those who don't drink?

(4) If you drive this type of car at a speed of 10 m/s to 15 m/s in the future, and the distance between you and the vehicle in front is kept between 42 m and 50 m, if you find that the vehicle in front stops suddenly, how many seconds should your reaction time be to prevent "rear-end"? Test center: the application of linear function. Special topic: Travel problem. Analysis: (1) substitute k=0. 1, t=0.5, v= 10 to get the braking distance;

(2) Substitute k=0. 1, v= 15 and s=20 into a given function to get the value of t;

(3) Substitute k=0. 1, v= 10 and t=2 into the given function to get the braking distance, and compare it with (1);

(4) Substitute k=0. 1, v= 15 and s=42 into the given function to get the value of t. Solution: (1) When k=0. 1, t=0.5 and v = 65433.

s = 0.5× 10+0. 1× 102 = 15m。

So the answer is:15;

(2)52.5 = 15t+0. 1× 152

The solution is t = 2.

So the answer is 2;

(3) When t=2 and v= 10, S = 2×10+0./kloc-0 /×102 = 30 (m).

30- 15= 15 (m).

Answer: The braking distance will be increased by15m compared with that without drinking;

(4) When v= 15 and s=42,