KINEMATICS O28

1. Analyze the Braking Process

We are given:

• Separation before braking ($S_{steady}$) = $110 \text{ m}$ (from Q26).
• Separation after stopping ($S_{final}$) = $10 \text{ m}$.
• Cars move at $v_{max} = 20 \text{ m/s}$ before braking.
• Car 2 brakes with a time delay $t_r$ after Car 1 brakes.
• Both decelerate at the same constant rate $a_b$.

2. Relate Separation Change to Delay

The reduction in separation is due to the fact that Car 2 continues moving at $v_{max}$ for time $t_r$ while Car 1 is already slowing down.
Consider the velocity-time graph of the braking phase. Car 2’s velocity profile is identical to Car 1’s but shifted by time $t_r$.

The distance traveled by Car 2 during braking phase is: $$ d_2 = (v_{max} \times t_r) + d_{brake} $$ The distance traveled by Car 1 during braking phase is: $$ d_1 = d_{brake} $$ (Where $d_{brake}$ is the distance covered during actual deceleration $v \to 0$).

The change in separation $\Delta S$ is the difference in distance traveled: $$ \Delta S = d_2 – d_1 = v_{max} \times t_r $$

3. Solve for Conditions on Deceleration ($a_b$)

We need the separation to drop from $110 \text{ m}$ to $10 \text{ m}$. $$ \Delta S = 110 – 10 = 100 \text{ m} $$ Using the relation derived above: $$ v_{max} \times t_r = 100 $$ $$ 20 \times t_r = 100 \implies t_r = 5 \text{ s} $$

The reaction time is fixed at $5 \text{ s}$ by the distance constraint. Does this constrain the deceleration $a_b$?
The only mathematical constraint provided in the problem statement is that the maximum deceleration capability is $4 \text{ m/s}^2$.
Therefore: $$ a_b \le 4 \text{ m/s}^2 $$ Within this limit, any constant deceleration will result in the same final separation of $10 \text{ m}$ as long as the delay is maintained at $5 \text{ s}$. The specific value of $a_b$ cancels out in the relative distance calculation ($d_2 – d_1$).