a boat is taking a cruise to a remote island 800 miles away. The trip downstream with current is 8 mile per hour. the return trip upstream against current is 10 hours. find the speed of the boat in still water and speed of current
v + c = 8
v - c on the way up
(v-c)10 = 800 so v - c = 80
NO WAY typo
Given:
d = 800 Miles.
T1 = 8 hours, downstream.
T2 = 10 hours, upstream.
(Vb+Vc)8 = 800.
Eq1: Vb + Vc = 100.
(Vb-Vc)10 = 800.
Eq2: Vb-Vc = 80.
Add Eq1 and Eq2:
Vb + Vc = 100.
Vb - Vc = 80.
2Vb = 180.
Vb = 90 mi/h.
In Eq1, replace Vb with 90:
90 + Vc = 100.
Vc = 10 mi/h.
To find the speed of the boat in still water and the speed of the current, we can set up a system of equations using the given information.
Let's assume the speed of the boat in still water is represented by "b" and the speed of the current is represented by "c."
1. We know that the trip downstream with the current took 8 hours to cover 800 miles. This can be represented as:
Distance = Speed × Time
800 = (b + c) × 8
2. We also know that the return trip upstream against the current took 10 hours to cover the same distance. This can be represented as:
Distance = Speed × Time
800 = (b - c) × 10
Now we have a system of equations to solve. We can use these equations to find the values of "b" and "c."
Let's start by rearranging the equations:
1. 800 = 8(b + c)
2. 800 = 10(b - c)
Now we can simplify them:
1. 100 = b + c
2. 80 = b - c
Now we can solve this system of equations. We have a couple of options, but one approach is to solve for "b" by adding equations 1 and 2:
(100 + 80) = (b + c) + (b - c)
180 = 2b
Divide both sides by 2:
b = 90
Substitute the value of "b" back into equation 1:
100 = 90 + c
c = 100 - 90
c = 10
Therefore, the speed of the boat in still water is 90 miles per hour, and the speed of the current is 10 miles per hour.