Chen's motorgoat took 3 hr to make a treip downstream with a 6-mph current. The return trip against the same current took 5 hr. Find the speed of the boat in still water.
since distance=speed*time, if the boat's speed is x, we have
3(x+6) = 5(x-6)
eq 1: y miles/(6 mph + x mph)=3 hr
eq 2: y miles/(6 mph - x mph)=5 r
multiply both sides by the numerator on the left
eq 1: y =18 + 3x
eq 2: y =30 - 5x
subtract equation 2 from equation 1
0=-12 + 8x
12=8x
12/8+x
To find the speed of the boat in still water, 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 represented by B (in mph).
On the downstream trip, the boat's effective speed is the sum of its speed in still water (B) and the speed of the current (6 mph). So, the effective speed is B + 6 mph. The distance covered in the downstream trip can be calculated by multiplying the effective speed by the time taken, which is 3 hours. Therefore, the distance is (B + 6) × 3.
On the return trip against the current, the boat's effective speed is the difference between its speed in still water (B) and the speed of the current (6 mph). So, the effective speed is B - 6 mph. The distance covered in the return trip can be calculated by multiplying the effective speed by the time taken, which is 5 hours. Therefore, the distance is (B - 6) × 5.
Since the distance traveled downstream and the distance traveled on the return trip are the same, we can set up the following equation:
(B + 6) × 3 = (B - 6) × 5
Now, let's solve this equation to find the value of B, which represents the speed of the boat in still water.
Expanding the equation, we get:
3B + 18 = 5B - 30
Grouping like terms, we have:
5B - 3B = 30 + 18
2B = 48
Dividing both sides of the equation by 2, we find:
B = 24
Therefore, the speed of the boat in still water is 24 mph.