I am having a hard time with this question.
A boat went 15 miles downstream in 1 hour and then returned upstream to its dock in 5/3 hours. if the current of the stream was constant, find the speed of the boat in still water and the speed of the current
speed of boat in still water --- x mph
speed of current ---------- y mph
so speed against the current = x-y mph
and speed with the current = x+y mph
downstream(with the current)
15/(x+y) = 1
x+y = 15
upstream:
15/(x-y) = 5/3
5x - 5y = 45
x - y = 9
add them:
2x = 24
x = 12, then since x+y=15, y = 3
speed of boat is 12 mph, the current is 3 mph
To solve this problem, you need to use the concept of relative speed. Let's assume that the speed of the boat in still water is 'b' miles per hour and the speed of the current is 'c' miles per hour.
When the boat is traveling downstream, the current helps it move faster. The effective speed of the boat is the sum of its speed in still water and the speed of the current, i.e., b + c miles per hour. So, using the given information, we can set up the equation:
15 = (b + c) * 1
Similarly, when the boat is traveling upstream, the current opposes its motion, so the effective speed is the difference between the boat's speed in still water and the speed of the current, i.e., b - c miles per hour. We can set up another equation using the given information:
15 = (b - c) * (5/3)
Now, we have two equations:
1. 15 = (b + c) * 1
2. 15 = (b - c) * (5/3)
To solve this system of equations, we can use one of the methods like substitution or elimination. Let's use the substitution method:
From equation 1, we can write b + c = 15.
Solving equation 2 for b - c, we get:
b - c = 15 * (3/5)
b - c = 9
Now, we can solve this system of equations:
b + c = 15
b - c = 9
Adding both equations, we get:
2b = 24
Dividing both sides by 2, we find:
b = 12
Substituting this value back into equation 1, we can solve for c:
12 + c = 15
c = 15 - 12
c = 3
Therefore, the speed of the boat in still water is 12 miles per hour, and the speed of the current is 3 miles per hour.