Two boat landings are 8.0 km apart on the same bank of a stream that flows at 2.8 km/h. A motorboat makes the round trip between the two landings in 50 minutes. What is the speed of the boat relative to the water?
I think it is something along the lines of
4/t+2.8 + 4/t-2.8 =5/6, but when I solve for x, I get the incorrect answer. Could you help me out?
To solve this problem, let's break it down step by step.
Let's assume the speed of the motorboat (relative to the water) is x km/h.
First, we need to find the time it takes for the boat to travel from one landing to the other. Since the distance between the landings is 8.0 km and the speed of the stream is given as 2.8 km/h, the boat's effective speed when going downstream (with the stream) would be (x + 2.8) km/h, and the time taken to travel downstream would be 8.0 km divided by (x + 2.8) km/h. Similarly, when going upstream (against the stream), the effective speed would be (x - 2.8) km/h, and the time taken would be 8.0 km divided by (x - 2.8) km/h.
According to the given information, the total round trip time is 50 minutes. Since the downstream and upstream journeys are equal distances, we can set up the equation:
8.0 / (x + 2.8) + 8.0 / (x - 2.8) = 50/60 (converting 50 minutes to hours)
Now, let's solve this equation to find the value of x:
Multiply both sides by (x + 2.8)(x - 2.8) to cancel out the denominators:
8.0(x - 2.8) + 8.0(x + 2.8) = (50/60)(x + 2.8)(x - 2.8)
Expand and simplify:
8x - 22.4 + 8x + 22.4 = (50/60)(x^2 - 2.8^2)
Combine like terms:
16x = (50/60)(x^2 - 2.8^2)
Multiply both sides by 60 to get rid of the fraction:
960x = 50(x^2 - 2.8^2)
Expand and simplify:
960x = 50x^2 - (50 * 2.8^2)
Rearrange the equation to bring it to standard quadratic form:
50x^2 - 960x - 50 * 2.8^2 = 0
Now, you can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 50, b = -960, and c = -50 * 2.8^2. Plugging these values into the formula, you can calculate x.
T = 50min * 1h/60min = 5/6 h.
T = t1 + t2 = 5/6 h.
d/(Vb-Vs) + d/(Vb+Vs) = T
8/(Vb-2.8) + 8/(Vb+2.8) = 5/6
LCD = (Vb-2.8)(Vb+2.8).
(8(Vb+2.8)+8(Vb-2.8))/(Vb-2.8)(Vb+2.8)=
5/6.
(8Vb+22.4+8Vb-22.4)/(Vb-2.8)(Vb+2.8)=5/6
16Vb/(Vb-2.8)(Vb+2.8) = 5/6
Cross multiply:
96Vb = 5(Vb-2.8)(Vb+2.8)
96Vb = 5(Vb^2-2.8^2)
96Vb = 5Vb^2-39.2
-5Vb^2 + 96Vb + 39.2 = 0
Use Quadratic formula and get:
Vb = 19.6 km/h = Speed of the boat.
Well, it seems like you're in quite a boatload of trouble with this question. But don't worry, I'm here to help you navigate through this mathematical stream with a touch of humor!
Let's break it down. The boat is traveling a total distance of 8.0 km, which includes going against the stream and coming back with the stream. So, the boat is effectively traveling 8.0 km going against the stream and 8.0 km with the stream.
Now, let's use our trusty formula: speed = distance / time. The boat's speed against the stream, let's call it v, is what we're looking for.
When the boat is going against the stream, its effective speed is the speed of the boat minus the stream's speed. So, the boat's speed against the stream is v - 2.8 km/h.
When the boat is coming back with the stream, its effective speed is the speed of the boat plus the stream's speed. So, the boat's speed with the stream is v + 2.8 km/h.
Now, let's use the formula for speed: speed = distance / time. The time it takes to go against the stream (8.0 km) is 8.0 km divided by the effective speed (v - 2.8 km/h). Similarly, the time it takes to come back with the stream is 8.0 km divided by the effective speed (v + 2.8 km/h).
Adding these times together, we get a total time of 50 minutes or 5/6 hours.
So, we can write the equation: (8.0 / (v - 2.8)) + (8.0 / (v + 2.8)) = 5/6.
Now, grab your favorite algebraic oar, and let's solve this equation to find the value of v!
Good luck, and remember to enjoy the ride!