Given Parameters

Part (a): Car A is Ahead

Scenario: Car A (lead) brakes hard immediately. Car B (follower) has a reaction time $t_d$ before braking gently.

Car A stops in distance $d_A = \frac{u^2}{2a_A} = \frac{u \cdot t_{sA}}{2}$.
Car B travels at constant velocity for $t_d$, then stops in $d_B = \frac{u \cdot t_{sB}}{2}$.
Total distance for B: $D_B = u t_d + \frac{u t_{sB}}{2}$.

For no collision, the initial gap $S_{min}$ must absorb the difference in stopping distances:

$$ S_{min} = D_B – d_A = u t_d + \frac{u}{2}(t_{sB} – t_{sA}) $$

Calculation:

$$ S_{min} = 30(1) + \frac{30}{2}(10 – 7) = 30 + 15(3) = 30 + 45 = 75 \text{ m} $$

Part (b): Car B is Ahead

Scenario: Car B (lead) brakes gently at $t=0$. Car A (follower) has reaction time $t_d$ then brakes hard.

Here, $a_A = u/t_{sA}$ and $a_B = u/t_{sB}$. Since $t_{sA} < t_{sB}$, the follower brakes harder ($a_A > a_B$). The danger of collision is only during the reaction time and the initial phase before A matches B’s velocity.

Derivation

The minimum separation $S_{min}$ corresponds to the “touch-and-go” case: the cars touch exactly when their velocities become equal. Let $t$ be the time when $v_A = v_B$.

$$ v_B(t) = u – a_B t \quad \text{and} \quad v_A(t) = u – a_A (t – t_d) $$

Equating velocities gives the time $t$:

$$ a_B t = a_A(t – t_d) \implies t = \frac{a_A t_d}{a_A – a_B} $$

Total minimum separation $S_{min}$ is the relative distance covered. It simplifies to:

$$ S_{min} = \frac{1}{2} a_B t_d^2 + \frac{a_B^2 t_d^2}{2(a_A – a_B)} $$

Substituting $a = u/t_s$ leads to the compact formula:

$$ S_{min} = \frac{t_d^2}{2} \cdot \frac{u}{t_{sB} – t_{sA}} $$

Calculation

Using $u = 30$ m/s, $t_d = 1.0$ s, $t_{sA} = 7.0$ s, $t_{sB} = 10.0$ s:

$$ S_{min} = \frac{30 \cdot (1.0)^2}{2(10.0 – 7.0)} = \frac{30}{2(3)} = \frac{30}{6} = 5.0 \text{ m} $$