Question
A car moves with a steady speed of
for 2 hours between two towns A and B. If the average speed of
the car for the round trip is
, then comptité the speed of the car when it moves from
to A.
for 2 hours between two towns A and B. If the average speed of
the car for the round trip is
, then comptité the speed of the car when it moves from
to A.
Answers
GPT-4o mini
To solve this problem, we need to clarify a few things and define the necessary variables.
1. Let's denote the speed of the car from Town A to Town B as \( v \).
2. The time taken to travel from Town A to Town B is 2 hours.
3. Therefore, the distance from A to B can be expressed as:
\[
d = v \times 2
\]
4. Let the speed of the car from Town B to Town A be \( u \).
5. 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:
\[
t = \frac{2v}{u}
\]
6. 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.
1. Let's denote the speed of the car from Town A to Town B as \( v \).
2. The time taken to travel from Town A to Town B is 2 hours.
3. Therefore, the distance from A to B can be expressed as:
\[
d = v \times 2
\]
4. Let the speed of the car from Town B to Town A be \( u \).
5. 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:
\[
t = \frac{2v}{u}
\]
6. 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.