A boat has a speed of 20 mph in still water. While traveling on a river at top speed, it went 40 miles upstream in the same amount of time it went 60 miles downstream. Find the rate of the river current.
since time=distance/speed,
40/(20-x) = 60/(20+x)
n still water, a boat averages 9 miles per hour. It takes the same amount of time to travel 26 miles downstream, with the current, as 10 miles upstream, against the current. What is the rate of the water's current?
n still water, a boat averages 9 miles per hour. It takes the same amount of time to travel 26 miles downstream, with the current, as 10 miles upstream, against the current. What is the rate of the water's current?
plug in the numbers 9 mi/hr, 26 mi, & 10 mi into the equation:
distance 1/(speed + x) = distance 2/(speed - x)
or
26/(9+x)=10/(9-x)
multiply both sides by (9-x)(9+x)
the denominators on either side can be cancelled, which then leaves the equation as
234-26x=90+10x
next, add/subtract like-terms
144=36x
x=4
To solve this problem, we can set up equations to represent the distance traveled by the boat both upstream and downstream.
Let's assume the rate of the river current is "c" mph.
When the boat travels upstream, it is moving against the current, so its effective speed is reduced. Therefore, its speed is (20 - c) mph.
When the boat travels downstream, it is moving with the current, so its effective speed is increased. Therefore, its speed is (20 + c) mph.
Now, let's use the formula: speed = distance/time, to set up the equations:
For the upstream journey:
Speed = distance / time
(20 - c) mph = 40 miles / t₁
For the downstream journey:
Speed = distance / time
(20 + c) mph = 60 miles / t₂
We are given that the time taken for both journeys is the same, so we can equate the two equations:
40 / t₁ = 60 / t₂
To simplify this equation, we can cross multiply:
40 * t₂ = 60 * t₁
Now, let's solve for t₁ and t₂:
t₁ = (40 * t₂) / 60
Now substitute this value of t₁ back into the upstream equation:
(20 - c) mph = 40 miles / ((40 * t₂) / 60)
Simplifying further:
(20 - c) mph = (40 * 60) / (40 * t₂)
20 - c = 60 / t₂
Now let's solve for t₂:
60 * t₁ = 40 * t₂
t₂ = (60 * t₁) / 40
Substitute this value of t₂ back into the equation:
20 - c = 60 / ((60 * t₁) / 40)
20 - c = 40 / t₁
Now, combining like terms:
20 - c = 40 / t₁
Multiply both sides by t₁:
20t₁ - c * t₁ = 40
Rearrange the equation:
c * t₁ = 20t₁ - 40
Rearrange again:
c * t₁ - 20t₁ = -40
Factor out t₁:
t₁(c - 20) = -40
Finally, solve for c:
c = -40 / (t₁ - 20)
Since we don't have the actual value of t₁, we cannot find the exact rate of the river current.