KINEMATICS BYU 3

Solution 3: Separation vs. Time Analysis

Step 1: Interpret the Separation ($s$) Graph

• $t=0$ to $t=10$: Constant separation $s=500$ m. Both cars are on the open road moving at $v_1$. Relative speed is 0.
• $t=10$ to $t=30$: Separation decreases linearly. This implies the front car has slowed down (entered the bridge with speed $v_2$) while the rear car is still fast ($v_1$).
• $t=30$ to $t=80$: Constant separation $s=200$ m. Both cars are on the bridge moving at $v_2$. Relative speed is 0.
• $t=80$ to $t=100$: Separation increases. The front car has exited the bridge (accelerated to $v_1$) while the rear car is still on the bridge ($v_2$).

Step 2: Calculate Relative Speed and $v_2$
During the interval $t \in [10, 30]$, the separation decreases from 500 m to 200 m. $$ \text{Slope} = \frac{\Delta s}{\Delta t} = \frac{500 – 200}{30 – 10} = \frac{300}{20} = 15 \text{ m/s} $$ This slope represents the relative speed: $v_{rel} = v_1 – v_2 = 15$ m/s.

During the constant intervals, separation distance is defined by velocity $\times$ time lag ($\Delta \tau$). Let $\Delta \tau$ be the constant time gap between the two cars passing any specific point. $$ s_{road} = v_1 \Delta \tau = 500 \text{ m} $$ $$ s_{bridge} = v_2 \Delta \tau = 200 \text{ m} $$ Divide the two equations: $$ \frac{v_1}{v_2} = \frac{500}{200} = 2.5 \implies v_1 = 2.5 v_2 $$ Substitute $v_1$ into the relative speed equation: $$ 2.5 v_2 – v_2 = 15 $$ $$ 1.5 v_2 = 15 \implies v_2 = 10 \text{ m/s} $$ Then: $$ v_1 = 2.5(10) = 25 \text{ m/s} $$

Step 3: Calculate Bridge Length
The front car is on the bridge from the moment it enters until the moment it leaves.

• Entry: The separation starts decreasing at $t=10$. So, Front Car enters at $t=10$.
• Exit: The separation starts increasing at $t=80$. So, Front Car exits at $t=80$.

Time spent on bridge $T_{bridge} = 80 – 10 = 70$ s. $$ \text{Length of bridge } L = v_2 \times T_{bridge} $$ $$ L = 10 \text{ m/s} \times 70 \text{ s} = 700 \text{ m} $$