A driver drives a car from area a to area B and arrives at the destination in 6 hours at an average speed of 80 km / h (1) When he returns on the original road, he finds the functional relationship between the time t (H) to reach Jiadi and the vehicle speed V (km / h) (2) If the driver returns at a constant speed, use 4.8h to calculate the return speed

A driver drives a car from area a to area B and arrives at the destination in 6 hours at an average speed of 80 km / h (1) When he returns on the original road, he finds the functional relationship between the time t (H) to reach Jiadi and the vehicle speed V (km / h) (2) If the driver returns at a constant speed, use 4.8h to calculate the return speed

1. The function relation is t = 480 / v
2. The return speed is 80 * 6 / 4.8 = 100km / h

The train route of a and B is 40 km longer than the bus route. The car starts from a with a speed of 60 km / h. after driving for 0.5 h, the train starts from a with a speed of 80 km / h. as a result, the car arrives at B one hour later than the train. How long is the train route and bus route of a and B

If the distance between the train and the car is x and y, then x = 40 + y
X △ 80 = 1.5 + y △ 60 because the car goes 0.5 hours first and arrives 1 hour later, so the time is 1.5 hours more
After solving x = 520 and y = 480, it takes 6.5 hours for the train and 8 hours for the car

The speed limit of a car in the center of a city is 40km / h. A woman was fined by the police for speeding, as shown in figure 2-3-5
Please use your physics knowledge to help the police explain the reason for being fined to this lady!
Excuse me for describing the picture in words:
Police: Madam, your speed was 60 km / h just now!
Woman: Sir, it's impossible. I just drove for seven minutes, but it's less than an hour. How can I walk 60 kilometers? What a joke!

Because the woman did not understand the physical meaning of speed, 60 km / h refers to the speed of the car, not the 7-minute journey. 60 km / h is greater than 40 km / h, so she was fined

The distance between a and B is 100 km, and a batch of goods and materials need to be delivered to B within 1 hour. Someone drives a car and walks 40 km in the first half hour, and the speed in the second half hour is at least 100 km
What's the minimum speed in the second half hour to ensure the timely delivery of this batch of materials?
Use the inequality of one variable to solve!

Let the velocity in the second half hour be a, then we can know from the known conditions:
0.5a+40=100
The solution is a = 120
That is, the speed in the second half hour is at least 120km / h

The distance between a and B is 100km. A person drives a car from a to B at a speed of 10m / s. half an hour after the person drives, another person drives a motorcycle from a to B. as a result, he arrives at the same time with the car and calculates the speed of the motorcycle

Half an hour away: 10M / s × 60s × 30 = 18000m
Remaining time: (100000-18000) m / (10m / s) = 8200s
Motorcycle speed v = 100000m / 8200s ≈ 12.20m/s
I hope my answer will help you, from the team of Baidu know you,

The vehicle runs for 100km at the speed of 100km / h, the engine power is 8kw, the efficiency is 25%, and the calorific value of known diesel engine is 3.3 × 10 ^ 7J / kg,
How many kilos of diesel fuel does this car consume when driving for 100km?

The driving time of the car is t = s / v = 100 / 100 = 1 hour,
Then the work of automobile engine is w = Pt = 8000 * 3600 = 2.88 × 10 ^ 7J,
The energy released by diesel in diesel engine is e = cm,
M = E / C = w / C, η = 2.88 × 10 ^ 7 / (3.3 × 10 ^ 7 × 25%) = 3.5kg

The speed of a car on a 100km road is 20m / s in the first 50km and 30m / s in the second 50km. What is the average speed on a 100km road

The average speed is
100000÷(50000÷20+50000÷30)
=2÷5/60
=24m/s

The distance between Party A and Party B is 1000 km. The logistics company undertakes the transportation task for the people. The speed of the car from Party A to Party B shall not exceed 80 km / h. It is known that the transportation cost of the car per hour consists of the variable part and the fixed part. The variable part is proportional to the square of the speed. The proportion coefficient is 1 / 100, and the fixed part is 54 yuan. If the whole transportation cost is 1500 yuan

(1000/X)*[(X^2/100)+54]=1500
X = 60, or x = 90 greater than 80, rounding off
The speed is 60

The distance between a and B is 100 km. What is the functional relationship between the speed y of the car from a to B and the time t? If the speed increases by 10 km / h, the time
How much less do you need

t=100/y
100/y-100/(y+10)=100(y+10-y)/y(y+10)=1000/(y^2+10y)

Is the average speed distance SKM a function of travel time th and why

Is a positive scaling function s = vt