Let's break down the problem and solve it step by step. **Given:** - Car A moves at a constant speed \( v_A = 30 \) m/s. - Car B starts from rest (initial velocity \( u_B = 0 \)), with constant acceleration \( a \), at the moment car A passes it. - We are to find: 1. The time (\( t \)) when car B catches up to car A 2. The distance car B travels during this time ### Step 1: Write the equations of motion for both cars Let’s choose the point where both cars start (when car A passes car B) as \( x = 0 \) and \( t = 0 \). **Car A:** Car A moves at constant speed: \[ x_A = v_A t = 30t \] **Car B:** Car B starts from rest with constant acceleration: \[ x_B = \frac{1}{2} a t^2 \] ### Step 2: Set the positions equal to each other (catch-up condition) At the time of catch-up, both cars are at the same position: \[ x_A = x_B \] \[ 30t = \frac{1}{2} a t^2 \] ### Step 3: Solve for \( t \) Rearrange the equation: \[ 30t = \frac{1}{2} a t^2 \] \[ 30t = 0.5 a t^2 \] Divide both sides by \( t \) (for \( t \neq 0 \)): \[ 30 = 0.5 a t \] \[ t = \frac{30}{0.5 a} \] \[ t = \frac{60}{a} \] ### Step 4: Find the distance car B traveled (use \( x_B \)) \[ x_B = \frac{1}{2} a t^2 \] Plug in \( t = \frac{60}{a} \): \[ x_B = \frac{1}{2} a \left(\frac{60}{a}\right)^2 \] \[ x_B = \frac{1}{2} a \cdot \frac{3600}{a^2} \] \[ x_B = \frac{1}{2} \cdot \frac{3600}{a} \] \[ x_B = \frac{1800}{a} \] --- ## **Final Answers** **A.** - At what time does car B catch up? \[ \boxed{t = \frac{60}{a}} \] - How far did it travel? \[ \boxed{x_B = \frac{1800}{a}} \] Where \( a \) is the constant acceleration of car B (in m/s²). If you know the value of \( a \), you can substitute it in for numerical values.