KINEMATICS BYU 1

Solution 1: Average Speed of the First Train

Let us define the variables:

• $s = 20$ km: Distance between stations.
• $v_p = 60$ km/h: Speed of the passenger’s train.
• $\Delta t_1 = 30 \text{ min} = 0.5$ h: Time interval the first train passed Station A before the passenger.
• $\Delta t_2 = 20 \text{ min} = \frac{1}{3}$ h: Time interval the first train arrived at Station B before the passenger.

Step 1: Analyze the Passenger’s Time
Let us set the time $t=0$ at the moment the passenger leaves Station A. The time taken by the passenger to reach Station B is: $$ t_p = \frac{s}{v_p} $$ Thus, the passenger arrives at Station B at absolute time $t = \frac{s}{v_p}$.

Step 2: Analyze the First Train’s Time
The problem states the first train passed Station A $\Delta t_1$ earlier than the passenger. $$ \text{Departure Time of Train} = -\Delta t_1 $$ The first train arrived at Station B $\Delta t_2$ earlier than the passenger. $$ \text{Arrival Time of Train} = t_p – \Delta t_2 = \frac{s}{v_p} – \Delta t_2 $$

Step 3: Calculate Total Duration for the First Train
The total time taken by the first train ($T_{train}$) is the difference between its arrival and departure times: $$ T_{train} = \text{Arrival} – \text{Departure} $$ $$ T_{train} = \left( \frac{s}{v_p} – \Delta t_2 \right) – (-\Delta t_1) $$ $$ T_{train} = \frac{s}{v_p} + \Delta t_1 – \Delta t_2 $$

Step 4: Determine Average Speed
The speed of the first train $v_{train}$ is distance divided by time: $$ v_{train} = \frac{s}{T_{train}} = \frac{s}{\frac{s}{v_p} + \Delta t_1 – \Delta t_2} $$ Multiply numerator and denominator by $v_p$ to simplify: $$ v_{train} = \frac{s v_p}{s + v_p(\Delta t_1 – \Delta t_2)} $$