Question

Два автомобиля выезжают одновременно из одного города в другой, находящийся на расстоянии 500км. Скорость первого на 10км/ч больше скорости второго, и поэтому первый автомобиль приезжает на место на 1ч раньше второго. Определите скорость каждого автомобиля!!!Умоляю, ребят, помогите)))

Answer 1

1*(х+10)+х=500
х+10+х=500
2х= 500-10
2х=480
х= 240

Answer 2

Problem Analysis

We have two cars traveling from one city to another, a distance of 500 km. The first car is traveling at a speed that is 10 km/h faster than the second car. The first car arrives at the destination 1 hour earlier than the second car. We need to determine the speed of each car.

Solution

Let's assume the speed of the second car is x km/h. Since the speed of the first car is 10 km/h faster, we can represent it as x + 10 km/h.

We know that the time taken by the first car to travel the distance is 1 hour less than the time taken by the second car. Let's represent the time taken by the second car as t hours. Therefore, the time taken by the first car is t - 1 hours.

We can use the formula speed = distance / time to calculate the speeds of the cars.

For the first car: speed of the first car = distance / time taken by the first car x + 10 = 500 / (t - 1)

For the second car: speed of the second car = distance / time taken by the second car x = 500 / t

We can solve these two equations to find the values of x and t.

Let's solve the equations:

x + 10 = 500 / (t - 1) (Equation 1) x = 500 / t (Equation 2)

We can solve Equation 1 for x and substitute it into Equation 2:

500 / t + 10 = 500 / (t - 1)

Now, let's solve this equation to find the value of t.

Calculation

To solve the equation 500 / t + 10 = 500 / (t - 1), we can cross-multiply and simplify:

500(t - 1) + 10t(t - 1) = 500t

Expanding and simplifying:

500t - 500 + 10t^2 - 10t = 500t

Rearranging the terms:

10t^2 - 10t - 500 = 0

Dividing the equation by 10:

t^2 - t - 50 = 0

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

Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -1, and c = -50.

Substituting the values into the quadratic formula:

t = (-(-1) ± √((-1)^2 - 4(1)(-50))) / (2(1))

Simplifying:

t = (1 ± √(1 + 200)) / 2

t = (1 ± √201) / 2

We have two possible values for t:

t1 = (1 + √201) / 2 t2 = (1 - √201) / 2

Since time cannot be negative, we can discard t2.

Now, let's substitute the value of t1 back into Equation 2 to find the value of x:

x = 500 / t1

Calculating x:

x = 500 / ((1 + √201) / 2)

Simplifying:

x = 1000 / (1 + √201)

Therefore, the speed of the second car is approximately 1000 / (1 + √201) km/h, and the speed of the first car is approximately 1000 / (1 + √201) + 10 km/h.

Answer

The speed of the second car is approximately 1000 / (1 + √201) km/h, and the speed of the first car is approximately 1000 / (1 + √201) + 10 km/h.

Please note that the values provided are approximate due to the use of the quadratic formula and the square root of an irrational number.