Given: - Speed of train A: \(v_A = 72 \, \text{km/h}\) - Speed of train B: \(v_B = 108 \, \text{km/h}\)
To convert the speeds from km/h to m/s:
\[ v_A = 72 \times \frac{1000}{3600} = 20 \, \text{m/s} \] \[ v_B = 108 \times \frac{1000}{3600} = 30 \, \text{m/s} \]
The relative velocity of train \(B\) with respect to train \(A\) is given by:
\[ v_{BA} = v_B - (-v_A) = v_B + v_A \]
Substituting the values:
\[ v_{BA} = 30 + 20 = 50 \, \text{m/s} \]
Since train \(B\) is moving towards the south and train \(A\) is moving towards the north, the relative velocity is considered negative:
\[ v_{BA} = -50 \, \text{m/s} \]
The velocity of the ground with respect to train \(B\) is simply the negative of the velocity of train \(B\) with respect to the ground:
\[ v_{\text{ground with respect to } B} = -v_B = -30 \, \text{m/s} \]
The velocity of train \(B\) with respect to \(A\) is \(-50 \, \text{m/s}\) and the velocity of the ground with respect to \(B\) is \(-30 \, \text{m/s}\).