On Monday afternoon, Andrew rowed his boat with current for 4.5 hours and covered 27 miles, stopping in the evening at a campground. On Tuesday morning he returned to his starting point against the current in 6.75 hours. Find the speed of the current and the rate at which Andrew rowed in still water. (I have problems solving word problems)
speed down = (v+c)
time down =4.5 = 27/(v+c)
speed up = (v-c)
time up = 6.75 = 27/(v-c)
so
6.75(v-c) = 4.5(v+c)
6.75 v -4.5 v = 4.5 c + 6.75 c
2.25 v = 11.25 c
v = 5 c
now back
6.75 = 27/(5c-c) = 27/4c
27 = 27c
c = 1
What is V+C
To solve this word problem, we can use the formula d = rt, where d is the distance, r is the rate or speed, and t is the time.
Let's assume the rate at which Andrew rows in still water is r mph, and the speed of the current is c mph.
According to the problem, Andrew rowed his boat with the current for 4.5 hours and covered 27 miles. We can set up the equation as follows:
27 = (r + c) * 4.5
Similarly, when Andrew rowed against the current, the distance covered is still 27 miles, but the time taken is 6.75 hours:
27 = (r - c) * 6.75
Now we have a system of equations:
1. (r + c) * 4.5 = 27
2. (r - c) * 6.75 = 27
To solve this system of equations, we can use the method of substitution or elimination.
Let's start by solving equation 2 for r:
(r - c) * 6.75 = 27
r - c = 4
We can rewrite this as r = c + 4 and substitute this into equation 1:
(c + 4 + c) * 4.5 = 27
(2c + 4) * 4.5 = 27
9c + 18 = 27
9c = 9
c = 1
Now that we have the value for c (speed of the current), we can substitute it back into equation 1 to find r:
(r + 1) * 4.5 = 27
4.5r + 4.5 = 27
4.5r = 22.5
r = 5
So, the speed of the current is 1 mph, and the rate at which Andrew rows in still water is 5 mph.