One day you were on a picnic with your class. During return journey from the picnic spot to your sch

Q: Why do we equate the actual time to the scheduled time?

The problem explicitly states the bus 'covered the remaining $s = 40 \text{ km}$ exactly in scheduled time.' This implies the total duration of the trip matched the originally planned duration, despite the varying speeds.

Q: Can I solve this using average speed formulas?

While technically possible, it's much more straightforward and less prone to error to use the basic relation $\text{Time} = \text{Distance} / \text{Speed}$ for each individual segment of the journey and sum them up.

Q: What if the bus didn't reach on time and was delayed?

If the bus was delayed by a specific time $\Delta t$, the master equation would simply be modified to: $t_1 + t_2 = T_0 + \Delta t$.

Q: Why did we convert the final answer to minutes?

The calculation gives the time in hours ($4/15 \text{ h}$), which is a fraction. Converting it to minutes ($16 \text{ min}$) provides a much more intuitive and standard way to express a short duration.

Q: Is knowing the total distance $D$ necessary to find the answer?

We introduce $D$ to set up the initial equations logically, but as you see in the algebraic steps, it gets substituted out. The final formula only depends on the given speeds and the remaining distance $s$.

Question

One day you were on a picnic with your class. During return journey from the picnic spot to your school, it began to rain, therefore the driver reduced speed of the bus and drove with an average speed $v_1 = 60 	ext{ km/h}$ instead of the scheduled average speed $v_0 = 70 	ext{ km/h}$. After the rain stopped, the driver drove the bus at an average speed $v_2 = 75 	ext{ km/h}$ and covered the remaining $s = 40 	ext{ km}$ exactly in scheduled time. How long did it rain?

Sigma Percentile · Accepted Answer

### The Picnic Downpour: A Kinematics Mystery

Imagine you are on a bus, heading back from an exhilarating class picnic. The driver is cruising steadily at a scheduled average speed of $70 	ext{ km/h}$. Everything is going exactly according to plan, and you expect to reach the school right on time.

But suddenly, the sky darkens, and a heavy downpour begins! For safety, the driver is forced to reduce the speed to $60 	ext{ km/h}$.

After some time, the rain finally stops. Realizing that the bus is now running behind schedule, the driver accelerates to $75 	ext{ km/h}$ for the remaining $40 	ext{ km}$ of the journey.

Miraculously, as the bus pulls into the school gates, you check your watch. You have arrived exactly on the originally scheduled time!

The question that naturally arises is: **Exactly how long did it rain?**

### Deconstructing the Journey

To solve this mystery, we must translate this physical scenario into the language of mathematics. Let's start by defining the overall parameters of the trip.

Let the total distance from the picnic spot to the school be $D$. If the rain had never occurred, the bus would have traveled this entire distance at the scheduled speed $v_0 = 70 	ext{ km/h}$.

Therefore, the originally scheduled time for the journey, which we will call $T_0$, is simply the total distance divided by the scheduled speed.

$$ T_0 = \frac{D}{v_0} $$

Now, let's analyze the actual journey, which is split into two distinct phases due to the weather.

**Phase 1: The Rainy Segment.** Let's assume the rain lasted for a time duration $t_1$. During this period, the bus traveled at a reduced speed $v_1 = 60 	ext{ km/h}$. The distance covered during this downpour is simply $v_1 t_1$.

**Phase 2: The Catch-up Segment.** After the rain stops, the bus has a remaining distance $s = 40 	ext{ km}$ to cover. The driver tackles this segment at an increased speed $v_2 = 75 	ext{ km/h}$. The time taken for this final stretch is $s / v_2$.

### The Master Equation

Here is the absolute crux of the problem. The problem explicitly states that the bus covered the remaining distance exactly in the scheduled time.

This means that despite the initial delay caused by the rain, the **total actual time taken** for the journey perfectly matches the **originally scheduled time**.

We can express this powerful constraint mathematically by equating the sum of the times of the two phases to the scheduled time $T_0$.

$$ t_1 + \frac{s}{v_2} = \frac{D}{v_0} $$

This is our master equation. However, we have an unknown variable $D$ (the total distance) that we need to eliminate.

We know that the total distance $D$ is simply the sum of the distance covered in the rain and the remaining distance $s$.

$$ D = v_1 t_1 + s $$

### Algebraic Symphony

Now, we substitute this expression for $D$ back into our master time equation. This will give us an equation where the only unknown is $t_1$, the duration of the rain.

$$ t_1 + \frac{s}{v_2} = \frac{v_1 t_1 + s}{v_0} $$

It is time for some careful algebraic manipulation. We want to isolate $t_1$. Let's split the fraction on the right side.

$$ t_1 + \frac{s}{v_2} = \frac{v_1}{v_0} t_1 + \frac{s}{v_0} $$

Next, we bring all the terms containing $t_1$ to the left side of the equation, and move the other terms to the right side.

$$ t_1 - \frac{v_1}{v_0} t_1 = \frac{s}{v_0} - \frac{s}{v_2} $$

Factoring out $t_1$ on the left side and $s$ on the right side reveals a beautiful, symmetric relationship.

$$ t_1 \left( \frac{v_0 - v_1}{v_0} ight) = s \left( \frac{v_2 - v_0}{v_0 v_2} ight) $$

Finally, by isolating $t_1$, we arrive at a clean and elegant formula. Notice how the scheduled speed $v_0$ in the denominators cancels out perfectly!

$$ t_1 = \frac{s(v_2 - v_0)}{v_2(v_0 - v_1)} $$

This formula is remarkable. It tells us that the duration of the rain depends **only** on the three speeds and the remaining distance. The total distance of the trip is completely irrelevant!

### The Final Calculation

With our symbolic formula ready, the final step is simply to plug in the numerical values provided in the problem.

We know the scheduled speed $v_0 = 70$, the rain speed $v_1 = 60$, the catch-up speed $v_2 = 75$, and the remaining distance $s = 40$.

$$ t_1 = \frac{40(75 - 70)}{75(70 - 60)} $$

Simplifying the terms inside the parentheses gives us a straightforward arithmetic calculation.

$$ t_1 = \frac{40 	imes 5}{75 	imes 10} $$

$$ t_1 = \frac{200}{750} = \frac{4}{15} 	ext{ hours} $$

While $4/15$ of an hour is mathematically correct, it is not a very intuitive way to express a short duration of time. Let's convert this value into minutes by multiplying by $60$.

$$ t_1 = \frac{4}{15} 	imes 60 	ext{ minutes} $$

$$ t_1 = 16 	ext{ minutes} $$

And there we have our answer! The sudden downpour that threatened to delay the class picnic lasted for exactly **$16$ minutes**.

By carefully breaking down the journey, setting up the time constraint, and trusting the algebra, we unraveled the kinematics mystery perfectly.

Visualized Solution

Journey Setup

Splitting the Trip

The Time Constraint

Substituting Distance

Algebraic Rearrangement

Solving for $t_{1}$

Numerical Substitution

Final Answer

The Way Forward

The Sigma Insight: Motion in a Straight Line

The Picnic Downpour

Deconstructing the Journey

The Master Equation

Algebraic Symphony

The Final Calculation

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