Моторная лодка прошла против течения реки 210 км и вернулась в пункт отправления,затратив на
обратный путь на 10 часов меньше,чем на путь против течения. Найдите скорость лодки в неподвижной воде, если скорость течения реки 10 км.чОтветы на вопрос
210/(х-10)-210/(х+10)=10
210/(х-10)-210/(х+10)-10=0 разделим на 10
21/(х-10)-21/(х+10)-1=0 умножим на (х-10)(х+10)
21(х+10)-21(х-10)-(х-10)(х+10)=0
21х+210-21х+210-(х²-100)=0
21х+210-21х+210-х²+100=0
-х²+520=0
х²=520
х=√520
х=2√130
Problem Analysis
We are given that a motorboat traveled 210 km against the current of a river and then returned to the starting point, spending 10 hours less on the return journey than on the journey against the current. We need to find the speed of the boat in still water, given that the speed of the river current is 10 km/h.Solution
Let's assume the speed of the boat in still water is x km/h. The speed of the current is given as 10 km/h.When the boat is traveling against the current, its effective speed is reduced by the speed of the current. So, the speed of the boat against the current is (x - 10) km/h.
When the boat is traveling with the current, its effective speed is increased by the speed of the current. So, the speed of the boat with the current is (x + 10) km/h.
We are given that the boat traveled 210 km against the current and spent 10 hours more on this journey than on the return journey. Let's denote the time taken for the journey against the current as t hours. The time taken for the return journey will be (t - 10) hours.
Using the formula distance = speed × time, we can set up the following equations:
1. For the journey against the current: - Distance = 210 km - Speed = (x - 10) km/h - Time = t hours
Therefore, we have the equation: 210 = (x - 10) × t.
2. For the return journey: - Distance = 210 km - Speed = (x + 10) km/h - Time = (t - 10) hours
Therefore, we have the equation: 210 = (x + 10) × (t - 10).
We can solve these two equations to find the value of x.
Solution Steps
1. Substitute the value of 210 for the distance in both equations. 2. Simplify the equations. 3. Solve the resulting system of equations to find the value of x.Let's solve the equations now.
Solution
1. Substitute the value of 210 for the distance in both equations: - Equation 1: 210 = (x - 10) × t - Equation 2: 210 = (x + 10) × (t - 10)2. Simplify the equations: - Equation 1: 210 = xt - 10t - Equation 2: 210 = xt + 10t - 10x - 100
3. Solve the resulting system of equations: - Subtract Equation 1 from Equation 2 to eliminate xt: - 210 - (xt + 10t - 10x - 100) = 0 - Simplify: 110 - xt - 10t + 10x = 0 - Rearrange: 10x - xt - 10t = -110 - Factor out x and t: x(10 - t) - 10t = -110 - Divide by (10 - t): x = (10t - 110) / (10 - t)
Now we have the value of x in terms of t. We can substitute this value into Equation 1 to solve for t.
4. Substitute the value of x into Equation 1: - 210 = (x - 10) × t - Substitute (10t - 110) / (10 - t) for x: - 210 = ((10t - 110) / (10 - t)) - 10) × t - Simplify: 210 = (10t - 110 - 10(10 - t)) × t - Simplify further: 210 = (10t - 110 - 100 + 10t) × t - Simplify further: 210 = (20t - 210) × t - Simplify further: 210 = 20t^2 - 210t
5. Rearrange the equation to form a quadratic equation: - 20t^2 - 210t - 210 = 0
Now we can solve this quadratic equation to find the value of t.
6. Solve the quadratic equation: - Use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a) - Substitute a = 20, b = -210, and c = -210 into the formula: - t = (-(-210) ± √((-210)^2 - 4(20)(-210))) / (2(20)) - Simplify: t = (210 ± √(44100 + 16800)) / 40 - Simplify further: t = (210 ± √60900) / 40 - Simplify further: t = (210 ± 310) / 40
We have two possible values for t. Let's calculate them separately.
7. Calculate the two possible values of t: - t1 = (210 + 310) / 40 = 520 / 40 = 13 - t2 = (210 - 310) / 40 = -100 / 40 = -2.5
Since time cannot be negative, we discard the value of t2 = -2.5.
8. Substitute the value of t1 = 13 into the equation for x: - x = (10t - 110) / (10 - t) - Substitute t = 13: - **x =
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