The distance between a and B is 80 km. Three hours after a bus leaves a place, a car starts from a place. Its speed is three times that of the bus. It is known that the car arrives at B place 20 minutes later than the bus. Find the speed of the two cars

The distance between a and B is 80 km. Three hours after a bus leaves a place, a car starts from a place. Its speed is three times that of the bus. It is known that the car arrives at B place 20 minutes later than the bus. Find the speed of the two cars

Suppose the speed of the bus is x km / h, then the speed of the car is 3x km / h. according to the meaning of the problem, we get 80x = 803x + 3 − 13, and the solution, we get x = 20. After testing, x = 20 is the root of the original equation, and it conforms to the meaning of the problem.. 3x = 60. Answer: the speed of the bus and the car are 20 km / h and 60 km / h respectively

For the sake of safety, the necessary distance should be kept between the vehicles driving on the highway. It is known that the maximum speed limit Vmax of an expressway is 120km / h. suppose that the front vehicle stops suddenly, and the driver of the rear vehicle finds this situation. The displacement from braking to the deceleration of the vehicle is 17m, and the resistance of the vehicle during braking is 0.5 times of the gravity of the vehicle, What is the minimum distance between vehicles on the expressway? (G is taken as 10m / S & # 178;)
The answer is 128M. The school will start in 10 days

Analysis: in the reaction time, the car does a uniform linear motion, so the safety distance between cars is equal to the sum of the displacement of uniform motion and the displacement of uniform deceleration linear motion
Because f = μ mg = ma, a = μ g = 0.5 * 10 = 5m / S ^ 2
From V ^ 2 = 2as1, the driving distance (braking distance) of the vehicle after braking action is obtained
s1=v^2/(2a)=33.3^2/(2*5)=111m——————————————（120km/h=33.3m/s)
The reaction distance is 17m
Therefore, the distance between cars on the highway is at least s = 111 + 17 + 128M

In order to be safe, the necessary distance must be kept between the vehicles driving on the highway. The maximum speed limit of a certain highway is known to be 120 km / h. suppose that the vehicle in front stops suddenly, and the driver in the rear finds this situation, and operates the brake, The response time t is 0.50 seconds. The acceleration is 4m / m2 when braking. How many meters is the distance between cars on the highway at least? (it's better to be more detailed, with the whole process of solving the problem)

V = 120 km / h = 33 1 / 3 M / s
When v = 0 m / s, it will take t = 33 and 1 / 3 △ 4 = 8 and 1 / 3 seconds
The safe distance is: 0.5 * 33 and 1 / 3 + 1 / 2 * 4 * (8 and 1 / 3) ^ 2 = 155.56m
A: the distance between cars on the highway is at least 155.56 meters

4. The requirements of a certain section of expressway for vehicle speed are: the minimum speed is 60km / h, the maximum speed is 120km / h, and the speed is Maomao
1. The requirements of a certain section of highway for vehicle speed are: the minimum speed is 60km / h, and the maximum speed is 120km / h. Maomao's father is driving on this section of highway, which is 200km away from the destination. At this time, it's just 9:00 in the morning. Question: when should the vehicle arrive at the destination

Calculate the lowest speed, calculate the highest speed: 200 / 60 = 3.33 hours, 200 / 120 = 1.67 hours, so 9:00 + 3.33 hours is about 12:20, 9:00 + 1.67 hours is about 10:40, so 10:40 to 12:20 is not illegal