Question

Расстояние между двумя городами 480км. с одного места к другому выехали одновременно два автомобиля . скорость первого на 20км\в год больше за скорость другого , поэтому он приехал в пункт назначения на 2 года раньше, чем другой ,Найдите скорость каждого автомобиля .

Answer 1

Х-скорость 2,время 480/х
х+20-скорость 1,время 480/(х+20)
480/х-480/(х+20)=2
2х(х+20)-480(х+20-х)=0
2х²+40х-9600=0
х²+20х-4800=0
х1+х2=-20 и х1*х2=-4800
х1=-80 не удов усл
х2=60км/ч-скорость2
60+20=80км/ч-скорость 1

Answer 2

Problem Analysis

We have two cars that start at the same time from one location and travel a distance of 480 km to reach their destination. The first car travels at a speed that is 20 km/h faster than the second car. The first car arrives at the destination 2 hours earlier than the second car. We need to find the speed of each car.

Solution

Let's assume the speed of the second car is x km/h. Since the first car is traveling 20 km/h faster, its speed will be x + 20 km/h.

We know that time equals distance divided by speed. The time taken by each car can be calculated using this formula.

For the first car: time = distance / speed time = 480 / (x + 20)

For the second car: time = distance / speed time = 480 / x

According to the problem, the first car arrives 2 hours earlier than the second car. So we can set up the following equation:

(480 / (x + 20)) - (480 / x) = 2

Now we can solve this equation to find the value of x, which represents the speed of the second car.

Calculation

Let's solve the equation:

(480 / (x + 20)) - (480 / x) = 2

Multiplying through by x(x + 20) to eliminate the denominators:

480x - 480(x + 20) = 2x(x + 20)

Expanding and simplifying:

480x - 480x - 9600 = 2x^2 + 40x

2x^2 + 40x - 9600 = 0

Dividing through by 2:

x^2 + 20x - 4800 = 0

Now we can solve this quadratic equation to find the value of x.

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = 1, b = 20, and c = -4800.

Substituting the values:

x = (-20 ± √(20^2 - 4 * 1 * -4800)) / (2 * 1)

Simplifying:

x = (-20 ± √(400 + 19200)) / 2

x = (-20 ± √19600) / 2

x = (-20 ± 140) / 2

Now we have two possible values for x:

x1 = (-20 + 140) / 2 = 120 / 2 = 60

x2 = (-20 - 140) / 2 = -160 / 2 = -80

Since speed cannot be negative, we discard the negative value of x.

Therefore, the speed of the second car is 60 km/h.

The speed of the first car is x + 20 = 60 + 20 = 80 km/h.

Answer

The speed of the first car is 80 km/h and the speed of the second car is 60 km/h.