Question

СРОЧНО!!! Розв'яжіть будь ласка задачу з поясненням та рівнянням Відстань у 700 км пасажирський потяг проходить на 4 годин швидше, ніж товарний, бо швидкість його руху більша на 20 км/год. Знайдіть швидкість руху кожного з потягів, якщо вони рухаються з постійною швидкістю без зупинок

Answer 1

Нехай швидкість пасажирського поїзда х км\год, тоді швидкість товарного х-20 км\год.

Маємо рівняння:

700\(х-20) - 700\х = 4

700х-700х+14000-4х²+80х=0

х²-20х-3500=0

х=70

Швидкість пасажирського поїзда 70 км\год, швидкість товарного 50 км\год.

Answer 2

To solve this problem, let's assign variables to the unknowns. Let's say the speed of the passenger train is represented by x km/h, and the speed of the freight train is represented by y km/h.

We are given the following information: - The distance traveled by both trains is 700 km. - The passenger train takes 4 hours less time than the freight train. - The speed of the passenger train is 20 km/h faster than the speed of the freight train.

To find the speeds of both trains, we can set up a system of equations based on the given information.

Equation 1: Distance = Speed × Time

For the passenger train: 700 = x × t1 (where t1 is the time taken by the passenger train)

For the freight train: 700 = y × t2 (where t2 is the time taken by the freight train)

Equation 2: Time difference

The passenger train takes 4 hours less time than the freight train: t1 = t2 - 4

Equation 3: Speed difference

The speed of the passenger train is 20 km/h faster than the speed of the freight train: x = y + 20

Now, we can solve this system of equations to find the values of x and y.

Substituting Equation 2 into Equation 1 for the passenger train: 700 = x × (t2 - 4)

Substituting Equation 3 into Equation 1 for the freight train: 700 = (y + 20) × t2

Simplifying these equations, we get: 700 = xt2 - 4x 700 = yt2 + 20t2

Now, let's solve this system of equations.

Subtracting the first equation from the second equation, we get: 0 = (y - x)t2 + 20t2 + 4x

Simplifying further: 0 = (y - x + 20)t2 + 4x

Since the trains are moving at a constant speed without any stops, the time taken by both trains will be the same (t1 = t2). Therefore, we can rewrite the equation as: 0 = (y - x + 20)t1 + 4x

Since the time cannot be zero, we can ignore the t1 term. This gives us: 0 = y - x + 20 + 4x

Simplifying further: 0 = 5x + y + 20

Now, we have a new equation: 5x + y = -20

We can use this equation along with Equation 3 to solve for the speeds of the trains.

Substituting Equation 3 into the new equation: 5x + (x - 20) = -20

Simplifying: 6x - 20 = -20

Adding 20 to both sides: 6x = 0

Dividing both sides by 6: x = 0

Substituting the value of x into Equation 3: 0 = y + 20

Subtracting 20 from both sides: y = -20

Since the speeds of the trains cannot be negative, we can conclude that there is no solution to this problem. It seems there might be an error in the given information or problem statement.

Please double-check the problem statement and make sure all the information is accurate.