the speed of the current in Puget Sound is 5 mph. A barge travels with the current 26 miles and returns in 10 and 2/3 hours. what is its speed in still water?
x=speed of boat in still water.
assuming the problem implies 26 miles each way.
to figure this out, I wrote this out in simple miles per hour terms. Example:
(hours/mile)*miles + (hours/mile)*miles = hours
so:
(1/(x+5))*26+(1/(x-5))*26=10.6
x+5 represents the speed of the boat traveling with the current
x-5 represents the speed of the boat traveling against the current
Hopefully this gets you far enough to come up with about 8 miles per hour.
Unclear. Is 10 2/3 hours the total time or just the return time?
To find the speed of the barge in still water, we can use the concept of relative velocity.
Let's represent the speed of the barge in still water as 'x' mph.
When the barge travels with the current, the effective speed is the sum of the speed in still water and the speed of the current. Therefore, the effective speed is 'x + 5' mph.
Given that the barge travels 26 miles with the current and returns in 10 and 2/3 hours, we can use the formula:
Distance = Speed × Time
Using this formula for the barge traveling with the current:
26 = (x + 5) × Time
Now, let's calculate the time it takes for the barge to return against the current.
The time taken for the return journey is given as 10 and 2/3 hours, which can be written as 32/3 hours.
Using the formula for the barge traveling against the current:
26 = (x - 5) × Time
Substituting the value of Time as 32/3, we get:
26 = (x - 5) × (32/3)
Now, we can solve this equation to find the value of x, which represents the speed of the barge in still water.
26 = (32/3) × (x - 5)
26 = (32/3) × (x - 5)
26 × 3 = 32(x - 5)
78 = 32x - 160
78 + 160 = 32x
238 = 32x
238/32 = x
x = 7.4375
Hence, the speed of the barge in still water is approximately 7.44 mph.
To find the speed of the barge in still water, we can use the concept of relative velocity. Let's assume the speed of the barge in still water is 'x' mph.
When the barge travels with the current, its effective speed is the sum of its own speed in still water and the speed of the current. So, the speed of the barge with the current is 'x + 5' mph.
When the barge travels against the current, its effective speed is the difference between its own speed in still water and the speed of the current. So, the speed of the barge against the current is 'x - 5' mph.
Now, we can use the formula: Distance = Speed * Time to solve the problem.
Given that the barge travels 26 miles with the current and returns in 10 and 2/3 hours, we can set up the following equations:
26 = (x + 5) * t1 --- (1)
26 = (x - 5) * t2 --- (2)
Here, t1 represents the time taken to travel with the current and t2 represents the time taken to travel against the current.
To find the value of 't1', we need to convert 10 and 2/3 hours to a mixed fraction, which equals 32/3 hours.
Substituting the values, we get:
26 = (x + 5) * 32/3
To simplify the equation, we can cross multiply:
26 * 3 = (x + 5) * 32
78 = 32x + 160
Subtracting 160 from both sides:
32x = -82
Dividing by 32 on both sides:
x = -2.5625
Since speed cannot be negative, we discard this solution.
Hence, there is no valid speed in still water that satisfies the given conditions.