in still water, a speed boat travels 5 times faster than the current of the river. If the speed boat can travel 48 miles upstream and then back in 5 hours, find the rate of the current.

so it goes (5c - c) upstream

and (5c + c) downstream

t = d / r

5 = (48 / 4c) + (48 / 6c)

5c = 12 + 8

Let's assume the speed of the current is "c" miles per hour, and the speed of the speedboat in still water is "s" miles per hour.

When the speedboat is traveling upstream against the current, its effective speed is reduced because it has to overcome the current's force. Therefore, the speedboat's speed relative to the ground is (s - c) miles per hour.

When the speedboat is traveling downstream with the current, its effective speed is increased because it receives assistance from the current. Therefore, the speedboat's speed relative to the ground is (s + c) miles per hour.

We are given that the speedboat can travel 48 miles upstream and then back in a total of 5 hours. This can be expressed as:

Distance upstream / Speed upstream + Distance downstream / Speed downstream = Total time

48 / (s - c) + 48 / (s + c) = 5

To solve for the rate of the current (c), we need to rearrange and solve this equation.

First, multiply both sides of the equation by the least common multiple of the two denominators, which is (s - c)(s + c).

48(s + c) + 48(s - c) = 5(s - c)(s + c)

48s + 48c + 48s - 48c = 5(s^2 - c^2)

96s = 5(s^2 - c^2)

96s = 5s^2 - 5c^2

Rearranging this equation:

5c^2 + 5s^2 - 96s = 0

Now, we need to use the information given that the speedboat travels 5 times faster than the current:

s = 5c

Plugging this into the equation:

5c^2 + 5(5c)^2 - 96(5c) = 0

5c^2 + 25c^2 - 480c = 0

30c^2 - 480c = 0

Dividing both sides by 30c:

c - 16c = 0

c(30c - 16) = 0

Setting each factor equal to zero:

c = 0 or 30c - 16 = 0

Since the speed of the current cannot be zero (as there must be a current in the river), we can solve for c:

30c - 16 = 0

30c = 16

c = 16/30

c ≈ 0.53 miles per hour

Therefore, the rate of the current is approximately 0.53 miles per hour.

To find the rate of the current, let's break down the problem into steps:

Step 1: Assign variables

Let's assign variables to the unknowns in the problem. Let:
- x be the speed of the speed boat in still water (in miles per hour)
- c be the speed of the current (in miles per hour)

Step 2: Understand the problem

From the problem statement, we can determine two conditions:
1. The speed boat travels 5 times faster than the current of the river. So, the speed of the speed boat is 5 times the speed of the current: x = 5c.
2. The speed boat takes 48 miles to travel upstream and then back in 5 hours.

Step 3: Calculate the time taken to travel upstream

To determine the time taken to travel 48 miles upstream, we divide the distance by the relative speed (speed of the boat minus the speed of the current):
Time upstream = Distance / Relative speed

Given that the speed boat travels 48 miles upstream, the relative speed is (x - c):
Time upstream = 48 / (x - c)

Step 4: Calculate the time taken to travel downstream

To determine the time taken to travel 48 miles downstream, we divide the distance by the relative speed (speed of the boat plus the speed of the current):
Time downstream = Distance / Relative speed

Given that the speed boat travels 48 miles downstream, the relative speed is (x + c):
Time downstream = 48 / (x + c)

Step 5: Set up the equation

The problem states that the time taken for the round trip is 5 hours. So, we can set up the equation:
Time upstream + Time downstream = 5 hours
48 / (x - c) + 48 / (x + c) = 5

Step 6: Solve the equation

Let's substitute the value of x from the first condition (x = 5c) into the equation from step 5.

48 / (5c - c) + 48 / (5c + c) = 5
48 / 4c + 48 / 6c = 5

Let's find a common denominator for the fractions:

(48 * 6 + 48 * 4) / (4c * 6 + 6c * 4) = 5
(288 + 192) / (24c + 24c) = 5
480 / 48c = 5

Now, we can solve for c:

480 = 5 * 48c
480 = 240c
c = 480 / 240
c = 2

Step 7: Answer the question

The rate of the current is 2 miles per hour.