Question

Мост длиной 475 м поезд проходит за 45 сек C такой же скоростьу он проходит мимо дежурного по станцииөза 20 сек Найдите скорось поезда Найдите длину поезда Лучше без обьеснений но с уравнением

Answer 1

(475+х)/45=х/20
20(475+х)=45*х
9500+20х=45х
45х-20х=9500
25х=9500
х=380м - длина поезда
380/20=(475+380)/45=19 м/с - скорость поезда

Answer 2

Problem Analysis

We are given the following information: - Length of the bridge: 475 m - Time taken by the train to cross the bridge: 45 seconds - Time taken by the train to pass a station master: 20 seconds

We need to find: - Speed of the train - Length of the train

Solution

Let's assume the speed of the train is v m/s and the length of the train is x m.

To find the speed of the train, we can use the formula: Speed = Distance / Time

The distance covered by the train to cross the bridge is the sum of the length of the train and the length of the bridge. So, the distance is x + 475 m.

The time taken by the train to cross the bridge is given as 45 seconds. Therefore, we have the equation: (x + 475) / v = 45 [[1]]

Similarly, the distance covered by the train to pass the station master is the length of the train. So, the distance is x m.

The time taken by the train to pass the station master is given as 20 seconds. Therefore, we have the equation: x / v = 20 [[2]]

We now have a system of two equations with two variables. We can solve these equations to find the values of v and x.

Solving the Equations

Let's solve the system of equations using substitution.

From equation [[2]], we can express x in terms of v: x = 20v [[3]]

Substituting equation [[3]] into equation [[1]], we get: (20v + 475) / v = 45

Simplifying the equation: 20v + 475 = 45v

Bringing all the v terms to one side: 45v - 20v = 475

Simplifying further: 25v = 475

Dividing both sides by 25: v = 19

Now that we have the value of v, we can substitute it back into equation [[3]] to find x: x = 20 * 19 = 380

Answer

Therefore, the speed of the train is 19 m/s and the length of the train is 380 m.

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