Physics Problem: Relative Motion Solution

This solution simplifies the problem by analyzing the motion of the Dog relative to the Boy. Instead of tracking two separate objects, we look at how the gap between them closes.

1. The Relative Graph Concept

We plot Relative Velocity (\(v_{rel}\)) versus Time.

• Initial Relative Velocity: \( u_{rel} = v_{dog} – v_{boy} = 0 – 2 = \mathbf{-2 \, m/s} \).
• Relative Acceleration: \( a_{rel} = a_{dog} – a_{boy} = 2 – 0 = \mathbf{2 \, m/s^2} \).

2. Step-by-Step Calculation

• Find the initial lag (The Negative Area):
  At \(t=0\), relative velocity is -2. Since acceleration is +2, it takes 1 second to reach 0 relative velocity. The distance “lost” during this time is the area of the small red triangle below the axis: $$ \text{Area}_{neg} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1 \text{ m} $$
• Set up the Catch-up Equation (The Positive Area):
  After 1 second, the dog moves faster than the boy. The graph forms a large triangle (blue) representing the distance gained. Let the duration of acceleration in the positive region be \(t_0\). By symmetry, deceleration is also \(t_0\).
  
  The slope is 2, so the peak height is \(2t_0\). $$ \text{Area}_{pos} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2t_0) \times (2t_0) = 2t_0^2 $$
• Solve for \(t_0\):
  The Net Displacement must equal the initial separation distance (199m). $$ \text{Area}_{pos} – \text{Area}_{neg} = 199 $$ $$ 2t_0^2 – 1 = 199 $$ $$ 2t_0^2 = 200 $$ $$ t_0^2 = 100 \implies t_0 = 10 \text{ s} $$
• Calculate Total Time:
  Total Time = (Initial Lag Time) + (Positive Motion Time) $$ t_{total} = 1 \text{s} + 2t_0 $$ $$ t_{total} = 1 + 2(10) = 21 \text{ s} $$