NEET 2025 Physics Motion in One Dimension MCQ Question

To solve the problem, we need to analyze the relative motion of the girl on the scooty and the buses traveling between cities X and Y.

Given Data:

• Girl's speed, $v_g = 60$ km/h
• A bus passes her every 30 minutes in the same direction.
• A bus passes her every 10 minutes in the opposite direction.

Let’s denote:

• The speed of the bus as $v_b$ km/h.
• The time period of the bus service as $T$ minutes.

Step 1: Analyze the buses going in the same direction (X to Y)

When a bus is moving in the same direction as the girl, we can calculate the relative speed as: $v_{\text{relative, same}} = v_b - v_g$

The girl sees a bus every 30 minutes; this means that the distance between two consecutive buses in the same direction can be calculated as: $\text{Distance} = v_{\text{relative, same}} \times \text{Time} = (v_b - 60) \times 0.5 \text{ hours}$

Since buses leave every $T$ minutes, the distance between two buses leaving in the same direction is: $\text{Distance} = v_b \times \frac{T}{60} \text{ hours}$

Setting these two distances equal gives us: $(v_b - 60) \times 0.5 = v_b \times \frac{T}{60}$

Step 2: Analyze the buses going in the opposite direction (Y to X)

For buses coming in the opposite direction, the relative speed is: $v_{\text{relative, opposite}} = v_b + v_g$

The girl sees a bus every 10 minutes; thus, the distance between two consecutive buses in the opposite direction is: $\text{Distance} = (v_b + 60) \times \frac{1}{6} \text{ hours}$

Setting this distance equal to the distance between buses: $(v_b + 60) \times \frac{1}{6} = v_b \times \frac{T}{60}$

Step 3: Solve the equations

We now have the following two equations:

1. From the same direction: $(v_b - 60) \times 0.5 = v_b \times \frac{T}{60}$
2. From the opposite direction: $(v_b + 60) \times \frac{1}{6} = v_b \times \frac{T}{60}$

Rearranging the first equation:

$(v_b - 60) \times 0.5 = v_b \times \frac{T}{60}$ $30(v_b - 60) = v_b T$ $30v_b - 1800 = v_b T$ $30v_b - v_b T = 1800$ $v_b (30 - T) = 1800$ $v_b = \frac{1800}{30 - T} \quad \text{(1)}$

Rearranging the second equation:

$(v_b + 60) \times \frac{1}{6} = v_b \times \frac{T}{60}$ $10(v_b + 60) = v_b T$ $10v_b + 600 = v_b T$ $10v_b - v_b T = -600$ $v_b (10 - T) = -600$ $v_b = \frac{-600}{10 - T} \quad \text{(2)}$

Step 4: Equate (1) and (2)

Setting the two expressions for $v_b$ equal to each other gives us: $\frac{1800}{30 - T} = \frac{-600}{10 - T}$

Cross-multiplying and solving for $T$: $1800(10 - T) = -600(30 - T)$ $18000 - 1800T = -18000 + 600T$ $18000 + 18000 = 1800T + 600T$ $36000 = 2400T$ $T = 15 \text{ minutes}$

Step 5: Solve for $v_b$

Substituting $T = 15$ minutes back into equation (1): $v_b = \frac{1800}{30 - 15} = \frac{1800}{15} = 120 \text{ km/h}$