Kinematics on a Triangular Track
1. Analyze the Track Geometry
The track is a triangle ABC with sides $a=400$ m, $b=300$ m, and $c=500$ m. Observing the lengths, we see that $300^2 + 400^2 = 90000 + 160000 = 250000 = 500^2$. Thus, the track is a right-angled triangle with the right angle at C.
Total perimeter length $L = 300 + 400 + 500 = 1200$ m.
2. Calculate Velocities
Both bikers start at A. They meet at B for the first time after $\Delta t_1 = 4$ min.
- Biker 1 (Clockwise): Travels path A $\to$ C $\to$ B.
Distance $d_1 = b + a = 300 + 400 = 700$ m.
Speed $v_1 = \frac{d_1}{\Delta t_1} = \frac{700}{4} = 175$ m/min. - Biker 2 (Anticlockwise): Travels path A $\to$ B directly.
Distance $d_2 = c = 500$ m.
Speed $v_2 = \frac{d_2}{\Delta t_1} = \frac{500}{4} = 125$ m/min.
3. Time for One Full Lap
Let $T_1$ and $T_2$ be the time taken by each biker to complete one full round of the track ($L=1200$ m).
4. Synchronizing at Point B
For the bikers to meet at B again, they must both arrive at B simultaneously. Since they started synchronized at A (which we can treat as $t=-\Delta t_1$ relative to the first meeting at B), and established a synchronized meeting at B at $t=0$, any future meeting at B must occur after integer numbers of full laps.
Let the time elapsed be $\Delta t$. For Biker 1 to be at B, $\Delta t = n_1 T_1$. For Biker 2 to be at B, $\Delta t = n_2 T_2$. We equate these times:
$$ n_1 \left( \frac{48}{7} \right) = n_2 \left( \frac{48}{5} \right) $$ $$ \frac{n_1}{7} = \frac{n_2}{5} $$The smallest integer solution is $n_1 = 7$ and $n_2 = 5$.
5. Final Calculation
Substituting back to find the time interval:
Answer: 48 min