Answer:
- Average Speed back home: 60 km/h
Step-by-step explanation:
- Let's denote the average speed for the 180 km journey (from the wine farm to Ceres) as [tex]s[/tex] km/h
- The return trip) is done at a speed 30 km/h slower than the outward journey.
- The time taken for the return journey is 1 hour more than the outbound trip.
- Calculate the time taken for each part using the formula
Time = distance/speed
Distance is the same = 180 km
- Outbound trip:
Speed = s
[tex]t_1 = \dfrac{{180}}{s}[/tex]
- Return trip
Speed = s -30
[tex]t_2 = \dfrac{{180}}{s-3}[/tex]
- Return trip takes 1 hour longer:
[tex]t_2 = t_1 + 1[/tex]
- Substituting in terms of distance and speed we get:
[tex]\dfrac{{180}}{s-3} = \dfrac{{180}}{s} + 1[/tex]
- We can solve this equation to get the speed S and then S-30, the average speed home:
[tex]\dfrac{180}{s}+1=\dfrac{180}{s-30} \quad \cdots [1][/tex] - Multiply throughout by LCM: [tex]s(s-30)[/tex]:
[tex]\dfrac{180}{s}s\left(s-30\right)+1\cdot \:s\left(s-30\right)=\dfrac{180}{s-30}s\left(s-30\right)[/tex]
** Simplify:
[tex]\dfrac{180}{s}s\left(s-30\right) = \quad 180\left(s-30\right) = 180s - 5400[/tex]
** Simplify:
[tex]1\cdot \:s\left(s-30\right)= s^2 - 30s[/tex]
** Simplify:
[tex]\dfrac{180}{s-30}s\left(s-30\right) = 180s[/tex]
- Equation [1] becomes:
[tex]180s-5400 + s^2 - 30s = 180s[/tex]
[tex]s^2+150s-5400=180s[/tex]
=> [tex]s^2-30s-5400=0[/tex]
- Solve using factorization
[tex]s^2-30s-5400= (s - 90)(s+60) = 0[/tex]
- The possible solutions are
[tex]s = 90 \;or\; s = -60[/tex]
- Ignoring the negative root we get:
[tex]s = 90 \;km/h[/tex]
- [tex]\text{Return speed =\;90 -\;30 = 60 \;km/h}[/tex]
So the average speed back home was 60 km/h
- **Return journey**:
- Distance = 180 km
- Speed = \(s - 30\) km/h (30 km/h slower)
- Time = Distance / Speed
- \(t_2 = \frac{{180}}{{s - 30}}\)
Given that the return journey took 1 hour longer, we have:
\[ t_2 = t_1 + 1 \]
Substitute the expressions for \(t_1\) and \(t_2\):
\[ \frac{{180}}{{s - 30}} = \frac{{180}}{{s}} + 1 \]
To solve for \(s\), let's cross-multiply and rearrange the equation:
\[ 180s = 180(s - 30) + s(s - 30) \]
Solving further:
\[ 180s = 180s - 5400 + s^2 - 30s \]
\[ s^2 - 30s - 5400 = 0 \]
Now let's solve the quadratic equation:
\[ s^2 - 30s - 5400 = (s - 90)(s + 60) = 0 \]
Since speed cannot be negative, we take the positive value:
\[ s = 90 \]
Therefore, the average speed at which she traveled back home was **90 km/h**¹².
Source: Conversation with Bing, 4/7/2024
(1) Solved A lady travels 180 km from her wine farm to Ceres. On - Chegg. https://www.chegg.com/homework-help/questions-and-answers/lady-travels-180-km-wine-farm-ceres-return-journey-travels-night-travels-30-km-h-slower-av-q108210446.
(2) a lady travels 180km from her wine farm to ceres. on the return journey .... https://brainly.in/question/40430658.
(3) she travels at night and travels 30km/h slower. She lost 1 hour on the .... https://brainly.in/question/50834485.
