A boat travels with the current at a speed of 10 miles per hour with respect to land, then against the same current at a speed of 6 miles per hour with respect to land. Find (a) the speed of the current, and (b) the speed of the boat in still water.
Let's denote the speed of the boat in still water as "b" and the speed of the current as "c".
(a) To find the current speed, we can use the concept of relative velocity. When the boat is traveling with the current, the effective speed is the sum of the boat's speed in still water and the speed of the current. Therefore, we have:
10 mph = b mph + c mph
When the boat is traveling against the current, the effective speed is the difference between the boat's speed in still water and the speed of the current. Therefore, we have:
6 mph = b mph - c mph
We now have a system of two equations:
10 = b + c ...(1)
6 = b - c ...(2)
To solve this system, we can add equations (1) and (2):
10 + 6 = (b + c) + (b - c)
16 = 2b
Dividing both sides by 2, we get:
b = 8 mph
Now that we know the speed of the boat in still water, we can substitute this value into equation (1) to solve for the speed of the current:
10 = 8 + c
Subtracting 8 from both sides, we have:
2 = c
Therefore, the speed of the current is 2 mph.
(b) The speed of the boat in still water is 8 mph.
To find the speed of the current and the speed of the boat in still water, we can use a system of equations.
Let's denote the speed of the boat in still water as "b" and the speed of the current as "c".
We know that when the boat travels with the current, its effective speed is 10 miles per hour, and when it travels against the current, its effective speed is 6 miles per hour.
(a) To find the speed of the current, we can set up the equation:
b + c = 10
This equation represents the boat's speed in still water (b) added to the speed of the current (c) equals the effective speed of 10 miles per hour when traveling with the current.
(b) To find the speed of the boat in still water, we can set up the equation:
b - c = 6
This equation represents the boat's speed in still water (b) minus the speed of the current (c) equals the effective speed of 6 miles per hour when traveling against the current.
Now, we have a system of equations:
b + c = 10
b - c = 6
We can solve this system of equations to find the values of b (speed of the boat in still water) and c (speed of the current).
To eliminate the variable "c", we can add the two equations:
(b + c) + (b - c) = 10 + 6
This simplifies to:
2b = 16
Dividing both sides of the equation by 2, we get:
b = 8
So, the speed of the boat in still water is 8 miles per hour.
To find the speed of the current, we can substitute the value of b into one of the original equations. Let's use the first equation:
b + c = 10
Replacing b with 8, we get:
8 + c = 10
Subtracting 8 from both sides of the equation, we find:
c = 2
Therefore, the speed of the current is 2 miles per hour.
In summary, (a) the speed of the current is 2 miles per hour, and (b) the speed of the boat in still water is 8 miles per hour.
let Vc be the current speed and Vb be the boat speed in still water.
Here is what you know:
Vb + Vc = 10 (downstream speed)
Vb - Vc = 6 (upstream speed)
Adding the two equations gives you
2 Vb = 16
Solve for Vb and then Vc