A 10 km long straight road connects two towns A and B. Two cyclists start simultaneously, one from t

Q: Why did we assume the first meeting is 2 km from town B and not town A?

Because the wind blows from A to B, the cyclist starting from A is faster. In the same amount of time, the faster cyclist covers more distance. Thus, the meeting point must be past the midpoint (5 km), making it closer to town B.

Q: How do we know the second meeting is 6 km from town B and not town A?

The phrasing 'first 2 km and then 6 km away from one of the towns' implies a single reference town for both measurements. Since the first meeting was 2 km from B, the second meeting must be 6 km from B.

Q: Why couldn't cyclist B be the one who rested at town A?

Cyclist A reaches town B much faster than B reaches A. If B rested at A, A would have already completed the return journey before B even started heading back, meaning they wouldn't meet on the road for the second time.

Q: Why do we use position equations instead of just relative velocity for the second meeting?

Because they don't start their return journeys at the same time! Cyclist A rests and starts later, while cyclist B starts immediately upon reaching A. Position equations handle this time delay perfectly.

Q: What if the wind was blowing from B to A instead?

The entire scenario would just mirror. Cyclist B would be the faster one, the first meeting would be 2 km from A, and cyclist B would rest at town A for 18.75 minutes.

Question

A 10 km long straight road connects two towns A and B. Two cyclists start simultaneously, one from town A and the other from town B. On reaching the opposite town a cyclist immediately returns to his starting town whereas the other cyclist takes some rest and then returns to his starting town. Both of them can ride at a speed 20 km/h in absence of wind but during their whole journey uniform wind from town A to B increases speed of a cyclist going with the wind by the same amount as it decreases the speed of the cyclist going against the wind. Both the cyclists meet twice, first 2 km and then 6 km away from one of the towns. In which town and for what period a cyclist rests.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a $10 	ext{ km}$ long straight road connecting two towns, A and B. We have two cyclists starting their journey simultaneously from opposite ends.

Let the distance between the towns be $d = 10 	ext{ km}$. Cyclist $C_A$ starts from town A, and cyclist $C_B$ starts from town B.

There is a uniform wind blowing from town A to town B. This wind will push the first cyclist, increasing his speed, but it will oppose the second cyclist, reducing his speed.

Let the wind speed be $w$. The speed of $C_A$ with the wind is $v_1 = 20 + w$. The speed of $C_B$ against the wind is $v_2 = 20 - w$.

### The First Meeting: Unlocking the Wind Speed

They meet for the first time at some point. Let's say this point is at a distance $x$ from town A.

Since they started together, the time taken by both to reach this point must be exactly the same. We can equate their times using the fundamental relation of distance over speed.

$$ \frac{x}{20+w} = \frac{10-x}{20-w} $$

Let's cross-multiply and solve for $x$. Notice how the $wx$ terms cancel out beautifully on both sides.

$$ x(20-w) = (10-x)(20+w) $$

$$ 20x - wx = 200 + 10w - 20x - wx $$

$$ 40x = 200 + 10w \implies x = 5 + 0.25w $$

There is a catch here. Since the wind speed $w$ is positive, $x$ must be strictly greater than $5 	ext{ km}$.

This means the meeting point is past the halfway mark, closer to town B. The problem states they meet $2 	ext{ km}$ from a town. **It has to be town B!**

If the meeting point is $2 	ext{ km}$ from B, then its distance from A is $x = 8 	ext{ km}$. Substituting this back, we can find the wind speed.

$$ 8 = 5 + 0.25w \implies 0.25w = 3 \implies w = 12 	ext{ km/h} $$

This gives us the actual speeds: $v_1 = 32 	ext{ km/h}$ for the first cyclist and $v_2 = 8 	ext{ km/h}$ for the second.

### The Race to the Opposite Towns

Now, let's see when they reach their respective opposite towns. Cyclist A is flying with the wind.

$$ t_{A1} = \frac{10}{32} = \frac{5}{16} 	ext{ h} $$

Cyclist B, fighting the wind, takes a grueling amount of time to reach town A.

$$ t_{B1} = \frac{10}{8} = \frac{20}{16} 	ext{ h} $$

### The Logical Deduction: Who Rests?

Notice the huge time difference. Cyclist A arrives way earlier.

If cyclist B were the one to rest at town A, cyclist A would have already completed his return trip before B even started!

Therefore, **it must be cyclist A who takes a rest at town B.**

### The Second Meeting: Finding the Rest Time

Let's set up the equations for their return journeys. Cyclist A rests for time $T_R$, then heads back against the wind at $8 	ext{ km/h}$.

We write his position from town A as a function of time $t$.

$$ x_A = 10 - 8\left(t - \left(\frac{5}{16} + T_Right)ight) $$

Cyclist B immediately turns around and rides with the wind at $32 	ext{ km/h}$. His position from town A is:

$$ x_B = 32\left(t - \frac{20}{16}ight) $$

The problem states the second meeting is $6 	ext{ km}$ from the same town, which is town B. This means the meeting point is $4 	ext{ km}$ from town A.

By setting cyclist B's position to $4$, we can easily find the exact time of this second meeting.

$$ 32\left(t - \frac{20}{16}ight) = 4 \implies t - \frac{5}{4} = \frac{1}{8} \implies t = \frac{22}{16} 	ext{ h} $$

### Final Calculation

Finally, we substitute this meeting time into cyclist A's position equation and set it to $4$.

$$ 10 - 8\left(\frac{22}{16} - \frac{5}{16} - T_Right) = 4 $$

$$ 6 = 8\left(\frac{17}{16} - T_Right) \implies \frac{3}{4} = \frac{17}{16} - T_R $$

Solving this gives us the rest time $T_R$.

$$ T_R = \frac{17}{16} - \frac{12}{16} = \frac{5}{16} 	ext{ h} $$

Multiplying by $60$, we get the final answer in minutes.

$$ T_R = \frac{5}{16} 	imes 60 = 18.75 	ext{ minutes} $$

**Cyclist A rests in Town B for 18.75 minutes.**

Visualized Solution

$d = 10$

$v_{1} = 20 + w, v_{2} = 20 - w$

$t_{1} = \frac{x}{20 + w} = \frac{10 - x}{20 - w}$

$x = 5 + 0.25 w$

$x > 5 ⟹ x = 8$

$w = 12$

$t_{A 1} = \frac{5}{16}, t_{B 1} = \frac{20}{16}$

$t_{A 1} ≪ t_{B 1}$

$x_{A} (t) = x_{B} (t)$

$x_{2} = 4 ⟹ t = \frac{22}{16}$

$T_{R} = \frac{5}{16}$

$T_{R} = 18.75$

The Sigma Insight: Relative Velocity

Analyzing the Setup

The First Meeting

The Race to the Opposite Towns

The Logical Deduction

The Second Meeting

Final Calculation

Similar Questions

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is

A boy crosses a river twice on a straight path at an angle $ϕ = 3 0^{\circ}$ with the downstream direction, first time in two minutes and second time in four minutes. If his speed relative to river current is $v_{b / w} = 3 m/s$ in both the attempts, find speed of the river current.

A river is flowing from West to East at a speed of $5 m/min$ . A man on the South bank of the river, capable of swimming at $10 m/min$ in still water, wants to swim across the river in the shortest time. He should swim in a direction

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=10

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v1​=20+w,v2​=20−w

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​=20+wx​=20−w10−x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=5+0.25w

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>5⟹x=8

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w=12

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tA1​=165​,tB1​=1620​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tA1​≪tB1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xA​(t)=xB​(t)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​=4⟹t=1622​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TR​=165​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TR​=18.75