Question

Велосипедист виїхав зі сталою швидкістю з пункта А в пункт Б, відстань між якими 105 км. На наступний день він відправився назад зі швидкістю на 7км.год більшою за попередню. По дорозі він зробив зупинку на 4 год, в результаті чого він затратив на зворотний шлях стільки ж часу скільки на шлях з А до Б . Знайти швидкість з Б до А.

Answer 1

Ответ:

17,5

Объяснение:

x - скорость велосипедиста из Б до А, км/ч.

105/(x-7)=105/x +4

105/(x-7)=(105+4x)/x

105x=(x-7)(105+4x)

105x=105x+4x²-735-28x

4x²-28x-735=0; D=784+11760=12544

x₁=(28-112)/8=-84/8=-10,5 - ответ не подходит по смыслу.

x₂=(28+112)/8=140/8=17,5 км/ч

Answer 2

Problem Analysis

We are given that a cyclist traveled from point A to point B with a constant speed, and the distance between the two points is 105 km. On the next day, the cyclist traveled back from point B to point A with a speed that is 7 km/h faster than the previous day. During the return trip, the cyclist made a 4-hour stop, which resulted in the same amount of time being spent on the return trip as on the trip from A to B. We need to find the speed from B to A.

Solution

Let's assume the speed of the cyclist from A to B is x km/h. Therefore, the speed from B to A would be (x + 7) km/h.

We know that the time taken for the trip from A to B is the same as the time taken for the return trip, after accounting for the 4-hour stop. Let's denote the time taken for the trip from A to B as t hours.

The time taken for the return trip would be t + 4 hours.

Using the formula speed = distance / time, we can set up the following equation for the trip from A to B:

x = 105 / t [[1]]

And for the return trip from B to A:

(x + 7) = 105 / (t + 4) [[2]]

To find the speed from B to A, we can solve equation [[2]] for x and substitute it into equation [[1]].

Let's solve equation [[2]] for x:

x + 7 = 105 / (t + 4)

x = 105 / (t + 4) - 7

Now, substitute this value of x into equation [[1]]:

105 / t = 105 / (t + 4) - 7

Simplifying the equation:

105(t + 4) = 105t - 7t(t + 4)

105t + 420 = 105t - 7t^2 - 28t

7t^2 + 28t + 420 = 0

Dividing the equation by 7:

t^2 + 4t + 60 = 0

Using the quadratic formula, we can solve for t:

t = (-4 ± √(4^2 - 4 * 1 * 60)) / (2 * 1)

t = (-4 ± √(16 - 240)) / 2

t = (-4 ± √(-224)) / 2

Since the value under the square root is negative, there are no real solutions for t. This means that there is no valid solution for this problem.

Therefore, we cannot find the speed from B to A given the information provided.