The Mad River flows at a rate of 3 km/h. In order for a boat to travel 78.2 km upriver and then return in a total of 8 hr, how fast must the boat travel in still water?
what formula would be used to solve this problem?
I used the formula of
V = D/T
T = D/V, you know that
78.2/(V + 3) + 78.2/(V - 3) = 8 hr.
Solve for V, the speed of the boat in still water.
Is the answer 10 km/h?
Your equation is correct, but when I solve it I got 20 km/h
(and 20 satisfies the equation, 10 does not)
oh okay thanks.
To solve this problem, you can use the formula for time:
T = D/V
Where:
T is the time taken to travel a distance D at a speed of V.
In this case, the boat is traveling upriver and then returning downstream. The boat is traveling with the current downstream, so the effective speed is the sum of the boat's speed in still water and the current's speed. However, when the boat is traveling upriver, it is going against the current, so the effective speed is the difference between the boat's speed in still water and the current's speed.
Let's solve the equation step by step:
1. The time taken to travel distance 78.2 km upstream is given by:
T1 = 78.2 / (V - 3)
2. The time taken to travel distance 78.2 km downstream is given by:
T2 = 78.2 / (V + 3)
3. The total time taken for the round trip is given as 8 hours:
T1 + T2 = 8
4. Substitute the values obtained in steps 1 and 2 into the equation from step 3:
78.2 / (V - 3) + 78.2 / (V + 3) = 8
Now, let's solve this equation step by step to find the speed of the boat in still water (V):
1. Multiply both sides of the equation by (V - 3)(V + 3) to eliminate the denominators:
78.2(V + 3) + 78.2(V - 3) = 8(V - 3)(V + 3)
2. Expand and simplify the equation:
78.2V + 234.6 + 78.2V - 234.6 = 8(V^2 - 9)
3. Combine like terms and simplify further:
156.4V = 8V^2 - 72
4. Rearrange the equation and set it equal to zero:
8V^2 - 156.4V - 72 = 0
5. Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:
V = (-b ± √(b^2 - 4ac)) / 2a
For the equation 8V^2 - 156.4V - 72 = 0, where a = 8, b = -156.4, and c = -72, plug the values into the quadratic formula to find the possible values of V.
6. Calculate the values of V using the quadratic formula:
V = (-(-156.4) ± √((-156.4)^2 - 4(8)(-72))) / 2(8)
This gives us two possible values for V.
7. Calculate the first possible value of V:
V1 = (156.4 + √(156.4^2 + 4(8)(72))) / 16
V1 ≈ 10.56 km/h
8. Calculate the second possible value of V:
V2 = (156.4 - √(156.4^2 + 4(8)(72))) / 16
V2 ≈ -0.68 km/h
Since speed cannot be negative in this context, we discard the negative value of V.
Therefore, the boat must travel at a speed of approximately 10.56 km/h in still water.
Yes, you are correct in using the formula V = D/T to solve this problem. In this case, V represents the speed of the boat in still water, D represents the distance traveled, and T represents the time taken.
To solve the problem, you correctly set up the equation:
78.2 / (V + 3) + 78.2 / (V - 3) = 8 hr
This equation represents the time taken for the boat to travel upriver and downriver. To find the speed of the boat in still water, you need to solve for V.
To do that, you can multiply the entire equation by common denominators to eliminate fractions:
78.2(V - 3) + 78.2(V + 3) = 8(V + 3)(V - 3)
78.2V - 234.6 + 78.2V + 234.6 = 8(V^2 - 9)
156.4V = 8V^2 - 72
0 = 8V^2 - 156.4V - 72
0 = 2V^2 - 39.1V - 18
Next, you can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, using the quadratic formula will be the most efficient method:
V = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the equation 2V^2 - 39.1V - 18 = 0, we get:
V = (-(-39.1) ± √((-39.1)^2 - 4 * 2 * (-18))) / (2 * 2)
V = (39.1 ± √(1522.81 + 144)) / 4
V = (39.1 ± √(1666.81)) / 4
Calculating the square root of 1666.81, we get:
V ≈ (39.1 ± 40.84) / 4
Simplifying further:
V ≈ (39.1 + 40.84) / 4 ≈ 19.935
V ≈ (39.1 - 40.84) / 4 ≈ -0.435
Since speed cannot be negative, we discard the negative value. Therefore, the speed of the boat in still water is approximately V ≈ 19.935 km/h.
So, the answer is not 10 km/h, but rather approximately 19.935 km/h.