A motorboat that travels with a speed of 20 km/hour in still water has traveled 36 km against the current and 22 km with the current, having spent 3 hours on the entire trip. Find the speed of the current of the river.
To find the speed of the current of the river, we need to set up a system of equations based on the given information.
Let's assume the speed of the current is "c" km/hour.
Against the current:
The speed of the boat relative to the water is reduced by the speed of the current, so the effective speed is (20 - c) km/hour. The time taken to travel 36 km against the current is given by distance/speed, which is 36/(20 - c) hours.
With the current:
The speed of the boat relative to the water is increased by the speed of the current, so the effective speed is (20 + c) km/hour. The time taken to travel 22 km with the current is given by distance/speed, which is 22/(20 + c) hours.
The total time taken for the trip is 3 hours, so we can write the equation:
36/(20 - c) + 22/(20 + c) = 3
Now, let's solve this equation to find the value of "c", the speed of the current.
To simplify the equation, we can multiply both sides by (20 - c)(20 + c) to get rid of the denominators:
36(20 + c) + 22(20 - c) = 3(20 - c)(20 + c)
Expanding and rearranging, we have:
720 + 36c + 440 - 22c = 3(400 - c^2)
Combining like terms:
1160 + 14c = 1200 - 3c^2
Rearranging again:
3c^2 + 14c - 40 = 0
Now we have a quadratic equation. To solve it, we can either factor or use the quadratic formula.
Factoring the equation, we have:
(3c - 4)(c + 10) = 0
Setting each factor to zero, we have two possible solutions:
1) 3c - 4 = 0, which gives c = 4/3
2) c + 10 = 0, which gives c = -10
Since the speed of the current cannot be negative, we discard the solution c = -10.
Therefore, the speed of the current of the river is 4/3 km/hour.
Let's assume the speed of the current is 'c' km/hour.
When the motorboat travels against the current, its effective speed is reduced by the speed of the current. Therefore, the speed of the motorboat relative to the river is (20 - c) km/hour.
When the motorboat travels with the current, its effective speed is increased by the speed of the current. Therefore, the speed of the motorboat relative to the river is (20 + c) km/hour.
We are given that the motorboat traveled 36 km against the current and 22 km with the current, and the total time spent on the trip is 3 hours.
Using the formula: distance = speed × time, we can set up two equations:
36 = (20 - c) × t1 (1)
22 = (20 + c) × t2 (2)
Since we know that the total time spent on the trip is 3 hours, we can also write:
t1 + t2 = 3 (3)
Now, let's solve the system of equations.
From equation (1):
36 = (20 - c) × t1
We can rewrite t1 in terms of c:
t1 = 36 / (20 - c) (4)
From equation (2):
22 = (20 + c) × t2
We can rewrite t2 in terms of c:
t2 = 22 / (20 + c) (5)
Substituting equations (4) and (5) into equation (3):
36 / (20 - c) + 22 / (20 + c) = 3
To simplify the equation, we can multiply through by the denominators:
36(20 + c) + 22(20 - c) = 3(20 - c)(20 + c)
Expanding and simplifying:
720 + 36c + 440 - 22c = 3(400 - c^2)
Combining like terms:
1160 + 14c = 1200 - 3c^2
Rearranging the equation:
3c^2 + 14c - 40 = 0
Now we can solve this quadratic equation. Factorizing, we get:
(3c - 4)(c + 10) = 0
Setting each factor equal to zero:
3c - 4 = 0 or c + 10 = 0
Solving for c:
3c = 4 or c = -10
c = 4/3 km/hour (speed of the current)
Since the speed of the current cannot be negative, we discard the solution c = -10.
Therefore, the speed of the current is 4/3 km/hour.
since time = distance/speed,
36/(20-s) + 22/(20+s) = 3