To solve this problem, we need to clarify a few things and define the necessary variables.
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Let's denote the speed of the car from Town A to Town B as \( v \).
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The time taken to travel from Town A to Town B is 2 hours.
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Therefore, the distance from A to B can be expressed as: \[ d = v \times 2 \]
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Let the speed of the car from Town B to Town A be \( u \).
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The distance from Town B to Town A is the same, so we can also express it as: \[ d = u \times t \] where \( t \) is the time taken to travel back from B to A.
Since both distances are equal, we can equate the two distance equations: \[ v \times 2 = u \times t \] From this equation, we can express \( t \) as: \[ t = \frac{2v}{u} \]
- The average speed for the entire round trip is given by: \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
The total distance for the round trip is: \[ \text{Total Distance} = d + d = 2d = 2 \times (v \times 2) = 4v \]
The total time for the round trip is: \[ \text{Total Time} = 2 + t = 2 + \frac{2v}{u} \]
Now we can set up the equation for average speed: \[ \text{Average Speed} = \frac{4v}{2 + \frac{2v}{u}} \]
We can denote the average speed of the round trip as \( s \), so: \[ s = \frac{4v}{2 + \frac{2v}{u}} \]
This equation allows us to express \( u \) in terms of \( v \) and \( s \): \[ s \left(2 + \frac{2v}{u}\right) = 4v \] Multiplying through by \( u \) gives: \[ s \cdot 2u + 2sv = 4vu \]
Rearranging gives: \[ 2su + 2sv = 4vu \]
Now, isolating \( u \): \[ 2su - 4vu = -2sv \] \[ u(2s - 4v) = -2sv \] \[ u = \frac{-2sv}{2s - 4v} \]
Now we can calculate the speeds if we are given values for \( s \) and \( v \). If you provide those speeds, I'll be able to solve for \( u \) as well.
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