Source: High school physics (Chinese)
Problem Sets:
Problem
Train A is moving forward at a speed $v_1$. The driver suddenly notices another train B on the same track at a distance $s$ ahead, moving in the same direction at a smaller constant speed $v_2$, where $v_2 < v_1$. The driver of train A immediately applies the brakes, causing a constant deceleration of magnitude $a$.
[Q1] To prevent a collision, the magnitude of the deceleration $a$ must satisfy the condition:
$$a \ge \frac{(v_1 - v_2)^2}{2s}$$To solve this problem, we analyze the motion of train A relative to train B. This simplifies the problem to a single object (A) moving towards a stationary target (B).
Let the frame of reference be fixed to train B. The initial velocity of train A relative to train B is:
$$v_{rel, i} = v_1 - v_2$$Since $v_1 > v_2$, train A is approaching train B.
The acceleration of train A relative to train B is:
$$a_{rel} = a_A - a_B$$Train A has a constant deceleration of magnitude $a$, so $a_A = -a$. Train B moves at a constant velocity, so $a_B = 0$.
$$a_{rel} = -a$$The initial separation distance is $s$. To avoid a collision, train A must reduce its relative velocity to zero before or exactly when it has covered the initial separation distance $s$. The minimum deceleration occurs at the critical condition where the final relative velocity becomes zero precisely after covering the distance $s$.
$$v_{rel, f} = 0$$We use the time-independent kinematic equation relating displacement, initial velocity, final velocity, and acceleration:
$$v_f^2 = v_i^2 + 2 a \Delta x$$Applying this to the relative motion:
$$v_{rel, f}^2 = v_{rel, i}^2 + 2 a_{rel} s$$Substituting the relative quantities into the equation:
$$0^2 = (v_1 - v_2)^2 + 2(-a)s$$Now, we solve for the deceleration $a$:
$$0 = (v_1 - v_2)^2 - 2as$$ $$2as = (v_1 - v_2)^2$$ $$a = \frac{(v_1 - v_2)^2}{2s}$$This value represents the minimum magnitude of deceleration required to prevent a collision. Therefore, the deceleration $a$ must be at least this large.