KINEMATICS BYU 4

Solution 4: Possible Constant Speeds

Step 1: Analyze the Phase Difference
Let the distance between signals be $s = 1.00$ km. Let the duration of Green and Red lights be $\tau = 30$ s. The signals are synchronized such that adjacent signals have opposite phases.

• Signal 0 ($x=0$): Green in $[0, \tau]$, Red in $[\tau, 2\tau]$, Green in $[2\tau, 3\tau] \dots$
• Signal 1 ($x=s$): Red in $[0, \tau]$, Green in $[\tau, 2\tau]$, Red in $[2\tau, 3\tau] \dots$
• Signal 2 ($x=2s$): Same phase as Signal 0.

Step 2: Condition for Non-stop Travel
Assume the vehicle enters the road at Signal 0 at $t=0$ (start of Green). To pass Signal 1 without stopping, the vehicle must arrive when Signal 1 is Green. Signal 1 turns Green at $t=\tau$ and stays Green until $t=2\tau$. Therefore, the time taken to travel distance $s$ ($t_1$) must satisfy: $$ t_1 = \tau \times (\text{odd integer}) $$ Why odd? If $t_1 < \tau$, we arrive at Signal 1 while it is still Red. If $t_1 = \tau$, we arrive exactly as it turns Green. Let us test the travel time $T$ between any two adjacent signals. If $T = \tau$, the car arrives at Signal 1 at $\tau$ (Green start). It arrives at Signal 2 at $2\tau$ (Signal 2 turns Green at $2\tau$). This works perfectly. If we travel slower, say $T = 3\tau$, the car arrives at Signal 1 at $3\tau$. Signal 1 is Green in $[3\tau, 4\tau]$. This also works. Generally, the travel time $T$ to cover distance $s$ must be an odd multiple of $\tau$: $$ T = (2n + 1)\tau \quad \text{where } n = 0, 1, 2, \dots $$

Step 3: Calculate Possible Speeds
Speed $v = \frac{s}{T}$. Substituting the condition for $T$: $$ v = \frac{s}{(2n + 1)\tau} $$ Given $s = 1 \text{ km} = 1000 \text{ m}$ and $\tau = 30 \text{ s}$. Note: The answer requires km/h, so let’s keep units consistent. $\tau = 30 \text{ s} = \frac{30}{3600} \text{ h} = \frac{1}{120} \text{ h}$. $$ v = \frac{1}{\frac{1}{120}(2n + 1)} = \frac{120}{2n + 1} \text{ km/h} $$