Question

Train A is moving along two parallel rail tracks towards north with speed 72 km/h and train B is moving towards south with speed 108 km/h.Velocity of train B with respect to A and velocity of ground with respect to B are (in m/s) :

• A. –30 and 50
• B. –50 and –30
• C. –50 and 30
• D. 50 and –30

Answer 1

The Correct Option is C

To solve the problem of finding the velocity of train B with respect to train A and the velocity of the ground with respect to train B, let's break it down step-by-step:

1. Convert the speeds from km/h to m/s:
   • Speed of train A, \( V_A = 72 \, \text{km/h} \). To convert km/h to m/s, use the conversion factor \( \frac{5}{18} \):
   • \(V_A = 72 \times \frac{5}{18} = 20 \, \text{m/s}\).
   • Speed of train B, \( V_B = 108 \, \text{km/h} \):
   • \(V_B = 108 \times \frac{5}{18} = 30 \, \text{m/s}\).
2. Calculate the velocity of train B with respect to train A:
   • The formula for velocity of one object with respect to another is \( V_{\text{relative}} = V_{\text{object}} - V_{\text{reference}} \).
   • Since train B is moving in the opposite direction to train A, the relative velocity is:
   • \(V_{BA} = V_B - (-V_A) = V_B + V_A = 30 + 20 = 50 \, \text{m/s}\)
   • However, since they are moving in opposite directions, the velocity of B with respect to A is actually in the opposite direction:
   • \(V_{BA} = -50 \, \text{m/s}\).
3. Calculate the velocity of the ground with respect to train B:
   • The ground is stationary, so its velocity relative to anything is 0 m/s. However, from train B's perspective, the ground seems to be moving in the opposite direction of B.
   • The velocity of the ground with respect to train B is:
   • \(V_{\text{ground, B}} = 0 - (-V_B) = V_B = 30 \, \text{m/s}\).

Therefore, the velocity of train B with respect to train A is \(-50 \, \text{m/s}\), and the velocity of the ground with respect to train B is \(30 \, \text{m/s}\).

Correct Answer: –50 and 30

Answer 2

Given: - Speed of train A: \(v_A = 72 \, \text{km/h}\) - Speed of train B: \(v_B = 108 \, \text{km/h}\)

Step 1: Converting Speeds to SI Units

To convert the speeds from km/h to m/s:

\[ v_A = 72 \times \frac{1000}{3600} = 20 \, \text{m/s} \] \[ v_B = 108 \times \frac{1000}{3600} = 30 \, \text{m/s} \]

Step 2: Calculating the Velocity of B with Respect to A

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:

\[ v_{BA} = -50 \, \text{m/s} \]

Step 3: Calculating the Velocity of the Ground with Respect to B

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:

\[ v_{\text{ground with respect to } B} = -v_B = -30 \, \text{m/s} \]

Conclusion:

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 \(-30 \, \text{m/s}\).