Solution to Question 20
1. Visualizing the Motion
Let’s define the coordinate system and variables based on the problem statement. The wind blows from Josh towards Mike (left to right).
2. Determining the Fly’s Effective Speed
The fly’s speed relative to the ground depends on the wind direction. Let $v_F$ be the air speed of the fly and $u$ be the wind speed.
- Forward Trip (Towards Mike): The fly moves with the wind.
$$v_1 = v_F + u = 30 + 10 = 40 \text{ km/h}$$
- Return Trip (Towards Josh): The fly moves against the wind.
$$v_2 = v_F – u = 30 – 10 = 20 \text{ km/h}$$
3. Logic: The Time Ratio
The fly goes back and forth repeatedly. Instead of summing an infinite geometric series, we can find the average speed of the fly over one complete “lap” (Josh $\to$ Mike $\to$ Josh). This average speed remains constant for every lap because the velocities are constant.
Let $t_1$ be the time the fly travels towards Mike, and $t_2$ be the time the fly travels back towards Josh during any single lap.
Consider the relative speeds of approach:
- When flying towards Mike, the relative speed of approach between the fly and Mike is $v_{rel,1} = v_1 + v_M$.
- When flying back towards Josh, the relative speed of approach between the fly and Josh is $v_{rel,2} = v_2 + v_J$.
Calculations for Time Ratio:
Let current separation be $x$.
Time forward: $t_1 = \frac{x}{v_1 + v_M} = \frac{x}{40 + 15} = \frac{x}{55}$
During $t_1$, the cyclists move closer by $(v_J + v_M)t_1 = 40 t_1$. The new separation is $x’ = x – 40(\frac{x}{55}) = x(1 – \frac{40}{55}) = x(\frac{15}{55})$.
Time backward: $t_2 = \frac{x’}{v_2 + v_J} = \frac{x’}{20 + 25} = \frac{x’}{45}$
Substituting $x’$:
$$ t_2 = \frac{1}{45} \cdot \left( x \frac{15}{55} \right) = \frac{x}{3 \times 55} $$
Now, compare $t_1$ and $t_2$:
$$ \frac{t_1}{t_2} = \frac{x / 55}{x / (3 \times 55)} = 3 $$
Thus, $t_1 = 3 t_2$. The fly spends 3 times longer flying with the wind than against it in every lap.
4. Calculating Average Velocity of the Fly
Since the time ratio is constant, the average speed of the fly ($\langle v \rangle$) over the entire journey is the time-weighted average of its speeds.
$$ \langle v \rangle = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2} $$
Substitute $t_1 = 3 t_2$:
$$ \langle v \rangle = \frac{v_1 (3 t_2) + v_2 (t_2)}{3 t_2 + t_2} = \frac{3 v_1 + v_2}{4} $$
Plugging in the values ($v_1 = 40, v_2 = 20$):
$$ \langle v \rangle = \frac{3(40) + 20}{4} = \frac{120 + 20}{4} = \frac{140}{4} = 35 \text{ km/h} $$
5. Final Calculation
Now we find the total time the cyclists travel until they meet, and use the fly’s average velocity to find the total distance flown.
Total Time ($T$):
$$ T = \frac{d}{v_J + v_M} = \frac{24}{25 + 15} = \frac{24}{40} = 0.6 \text{ hours} $$
Total Distance Flown ($s$):
$$ s = \langle v \rangle \times T $$
$$ s = 35 \text{ km/h} \times 0.6 \text{ h} = 21 \text{ km} $$
Answer: The total distance flown by the fly is 21 km.
Let’s define the coordinate system and variables based on the problem statement. The wind blows from Josh towards Mike (left to right).
The fly’s speed relative to the ground depends on the wind direction. Let $v_F$ be the air speed of the fly and $u$ be the wind speed.
- Forward Trip (Towards Mike): The fly moves with the wind. $$v_1 = v_F + u = 30 + 10 = 40 \text{ km/h}$$
- Return Trip (Towards Josh): The fly moves against the wind. $$v_2 = v_F – u = 30 – 10 = 20 \text{ km/h}$$
The fly goes back and forth repeatedly. Instead of summing an infinite geometric series, we can find the average speed of the fly over one complete “lap” (Josh $\to$ Mike $\to$ Josh). This average speed remains constant for every lap because the velocities are constant.
Let $t_1$ be the time the fly travels towards Mike, and $t_2$ be the time the fly travels back towards Josh during any single lap.
Consider the relative speeds of approach:
- When flying towards Mike, the relative speed of approach between the fly and Mike is $v_{rel,1} = v_1 + v_M$.
- When flying back towards Josh, the relative speed of approach between the fly and Josh is $v_{rel,2} = v_2 + v_J$.
Calculations for Time Ratio:
Let current separation be $x$.
Time forward: $t_1 = \frac{x}{v_1 + v_M} = \frac{x}{40 + 15} = \frac{x}{55}$
During $t_1$, the cyclists move closer by $(v_J + v_M)t_1 = 40 t_1$. The new separation is $x’ = x – 40(\frac{x}{55}) = x(1 – \frac{40}{55}) = x(\frac{15}{55})$.
Time backward: $t_2 = \frac{x’}{v_2 + v_J} = \frac{x’}{20 + 25} = \frac{x’}{45}$
Substituting $x’$:
$$ t_2 = \frac{1}{45} \cdot \left( x \frac{15}{55} \right) = \frac{x}{3 \times 55} $$Now, compare $t_1$ and $t_2$:
$$ \frac{t_1}{t_2} = \frac{x / 55}{x / (3 \times 55)} = 3 $$Thus, $t_1 = 3 t_2$. The fly spends 3 times longer flying with the wind than against it in every lap.
Since the time ratio is constant, the average speed of the fly ($\langle v \rangle$) over the entire journey is the time-weighted average of its speeds.
$$ \langle v \rangle = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2} $$Substitute $t_1 = 3 t_2$:
$$ \langle v \rangle = \frac{v_1 (3 t_2) + v_2 (t_2)}{3 t_2 + t_2} = \frac{3 v_1 + v_2}{4} $$Plugging in the values ($v_1 = 40, v_2 = 20$):
$$ \langle v \rangle = \frac{3(40) + 20}{4} = \frac{120 + 20}{4} = \frac{140}{4} = 35 \text{ km/h} $$Now we find the total time the cyclists travel until they meet, and use the fly’s average velocity to find the total distance flown.
Total Time ($T$):
$$ T = \frac{d}{v_J + v_M} = \frac{24}{25 + 15} = \frac{24}{40} = 0.6 \text{ hours} $$Total Distance Flown ($s$):
$$ s = \langle v \rangle \times T $$ $$ s = 35 \text{ km/h} \times 0.6 \text{ h} = 21 \text{ km} $$