Question

Two cars P and Q are travelling on a straight path and are 60 m apart. Car P moves with \(36~\text{km/h}\), car Q moves with \(18~\text{km/h}\). At \(t=0\), driver in car P applies brakes, uniform deceleration occurs, and collision happens after \(30~\text{s}\). Find the correct velocity (in m/s) of car P just before collision.

• A. 1
• B. 16
• C. 5
• D. 4

Hint:

In collision/meeting problems with uniform acceleration, always use relative motion: \(\Delta x = v_{r,\text{avg}} \cdot t\). This avoids writing separate equations for both objects.

Answer 1

The Correct Option is D

Step 1: Convert given speeds to SI units.
\[ v_{P0} = 36 \times \frac{1000}{3600} = 10~\text{m/s}, v_{Q} = 18 \times \frac{1000}{3600} = 5~\text{m/s}. \] So initially: Car P = \(10~\text{m/s}\), Car Q = \(5~\text{m/s}\).

Step 2: Initial relative speed.
Relative speed at \(t=0\): \[ v_{r0} = v_{P0} - v_Q = 10 - 5 = 5~\text{m/s}. \]

Step 3: Condition for collision.
Initial separation = \(60~\text{m}\). Collision occurs in \(t=30~\text{s}\). Thus the relative displacement closed = \(60~\text{m}\). Hence, average relative speed is: \[ v_{r,\text{avg}} = \frac{\Delta x}{t} = \frac{60}{30} = 2~\text{m/s}. \]

Step 4: Use average of initial and final relative speeds.
For uniformly accelerated (or decelerated) motion: \[ v_{r,\text{avg}} = \frac{v_{r0} + v_{r,final}}{2}. \] Substitute values: \[ 2 = \frac{5 + v_{r,final}}{2}. \] \[ v_{r,final} = -1~\text{m/s}. \]



Step 5: Interpret final relative velocity.
\[ v_{r,final} = v_P' - v_Q = -1. \] So, \[ v_P' - 5 = -1 \Rightarrow v_P' = 4~\text{m/s}. \] This is the velocity of car P just before collision. Final Answer:
\[ \boxed{4~\text{m/s}} \]