Let the speed of the car on the circular path be $v_c$ and the speed of the car on the straight road be $v_s$. Since the problem states the relative velocity becomes zero at $t=25$ s, both cars must be moving in the same direction with the same speed at that instant.

$$ |\vec{v}_{rel}| = |\vec{v}_c – \vec{v}_s| = 0 \implies \vec{v}_c = \vec{v}_s $$

Thus, their speeds are equal: $v_c = v_s = v$.

Consider a small time interval $dt$ near the instant $t=25$ s. The velocity of the car on the straight road, $\vec{v}_s$, is constant. Therefore, change in its velocity $d\vec{v}_s = 0$.

The velocity of the car on the circular path, $\vec{v}_c$, changes direction. The magnitude of this change for a small interval is related to centripetal acceleration: $$ |d\vec{v}_c| = a_c \, dt = \frac{v^2}{R} \, dt $$

The change in relative velocity $d\vec{v}_{rel}$ is given by: $$ d\vec{v}_{rel} = d\vec{v}_c – d\vec{v}_s $$ $$ d\vec{v}_{rel} = d\vec{v}_c – 0 $$

Taking the modulus: $$ |d\vec{v}_{rel}| = |d\vec{v}_c| = \frac{v^2}{R} \, dt $$ $$ \frac{|d\vec{v}_{rel}|}{dt} = \frac{v^2}{R} $$

This implies that the slope of the $v_{rel}$ vs. time graph (near the zero crossing) represents the centripetal acceleration of the car.

We take the time interval from $t=25$ s to $t=30$ s (or from $t=20$ to $t=25$) where the graph is linear. Using the coordinates from the graph:

• At $t_1 = 25$ s, $v_{rel} = 0$ m/s.
• At $t_2 = 20$ s, $v_{rel} = 10$ m/s.

Calculating the magnitude of the slope: $$ \text{Slope} = \left| \frac{\Delta v}{\Delta t} \right| = \left| \frac{10 – 0}{20 – 25} \right| = \frac{10}{5} = 2\text{m/s}^2 $$

Equating this to the centripetal acceleration derived in Step 2: $$ \frac{v^2}{R} = 2$$

Given $R = 200$ m: $$ v^2 = 2\times 200 $$ $$ v^2 = 400 $$ $$ v = \sqrt{400} = 20 \text{m/s} $$