A motorvoat goes 36 miles upstream on a river whose current is running at 3 miles per hour. The trip up and back takes 5 hours. What is the speed of the boat ( assuming that it maintains a constant speed relative to the water)?
t going upstream
(5 - t) going downstream
speed upstream = (v-3)
speed downstream = (v+3)
36 = t(v-3)
36 = (5-t)(v+3)
36 = t v - 3 t
t = 36/(v-3)
36 = [5 - 36/(v-3)] (v+3)
36 = 5(v+3) - 36(v+3)/(v-3)
(21 - 5 v)(v-3) = -36(v+3)
check, multiply out and solve quadratic for v
To find the speed of the boat, we can set up a system of equations based on the given information.
Let's assume the speed of the boat in still water is "B" miles per hour.
When the boat is going upstream, against the current, the effective speed of the boat is reduced by the speed of the current. So the speed of the boat going upstream is (B - 3) miles per hour.
Similarly, when the boat is going downstream, with the current, the effective speed of the boat is increased by the speed of the current. So the speed of the boat going downstream is (B + 3) miles per hour.
Now, let's use the given information to form our equations:
1. When the boat goes upstream (against the current), it travels 36 miles. The time taken for this trip can be calculated using the formula: time = distance/speed. So, the time taken to go upstream is: 36/(B - 3) hours.
2. When the boat goes downstream (with the current), it covers the same distance of 36 miles. The time taken for this trip is: 36/(B + 3) hours.
According to the given information, the total time taken for the round trip is 5 hours:
Time upstream + Time downstream = 5
Plugging in the previously calculated times:
36/(B - 3) + 36/(B + 3) = 5
Now, we can solve this equation to find the value of "B" which represents the speed of the boat in still water.
Note: The equation has variables on both sides, so it cannot be solved algebraically. We can use numerical methods or trial and error to find the approximate value of "B" that satisfies the equation.