rowing. Abby rows 10 km upstream and 10 km back in a total time of 3 hours. The speed of the river is 5km/h.Find Abby's speed in still water.
10/(v-5)+10/(v+5)=3
10(v+5)+10(v-5)=3(v+5)(v-5)
20v=3v^2-75
3v^2-20v-75=0
To find Abby's speed in still water, we need to use the concept of relative motion and apply it to the given information.
Let's consider Abby's speed in still water as "S" km/h.
When Abby rows upstream against the river's flow, her effective speed is reduced by the speed of the river. So, her effective speed upstream is (S - 5) km/h.
Similarly, when Abby rows downstream, her effective speed is enhanced by the speed of the river. So, her effective speed downstream is (S + 5) km/h.
Now, let's calculate the time it took for Abby to row upstream and downstream.
Time taken to row upstream = Distance / Speed = 10 km / (S - 5) km/h
Time taken to row downstream = Distance / Speed = 10 km / (S + 5) km/h
According to the given information, the total time taken (upstream + downstream) is 3 hours.
Therefore, we can set up the equation:
10 / (S - 5) + 10 / (S + 5) = 3
To solve this equation for S, we need to first find a common denominator and then simplify the equation.
Multiplying through by (S - 5)(S + 5), we get:
10(S + 5) + 10(S - 5) = 3(S - 5)(S + 5)
Simplifying further:
10S + 50 + 10S - 50 = 3(S^2 - 25)
Combining like terms, we have:
20S = 3S^2 - 75
Rearranging the equation to bring it to standard quadratic form:
3S^2 - 20S - 75 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Once we solve for the value(s) of S, we can determine Abby's speed in still water.
Let's solve this equation using the quadratic formula:
S = (-b +/- sqrt(b^2 - 4ac)) / 2a
For our equation, a = 3, b = -20, and c = -75.
Plugging these values into the quadratic formula, we get:
S = (-(-20) +/- sqrt((-20)^2 - 4 * 3 * -75)) / (2 * 3)
= (20 +/- sqrt(400 + 900)) / 6
= (20 +/- sqrt(1300)) / 6
Now, calculating the two possible values for Abby's speed in still water:
S1 = (20 + sqrt(1300)) / 6
S2 = (20 - sqrt(1300)) / 6
So, Abby's speed in still water can be either (20 + sqrt(1300)) / 6 km/h or (20 - sqrt(1300)) / 6 km/h, depending on the solution of the quadratic equation.