Ashwin fires two bullets from the same place at an interval of 15 minutes. Rahul is sitting in a bus that is approaching the firing point. Rahul hears the second gunshot sound 14 minutes 30 seconds after he hears the first one. If the speed of sound in air is taken as 330 m/s, what is the approximate speed of the bus in metres per second?

Introduction / Context:
This question combines relative motion and the finite speed of sound. Two gunshots are fired 15 minutes apart, but because Rahul is moving toward the source, he hears the second sound only 14 minutes 30 seconds after the first. Using the difference in heard interval and the speed of sound, we must determine the approximate speed of the bus.

Given Data / Assumptions:

• Time interval between gunshots (actual firing times) = 15 minutes.
• Time interval between Rahul hearing the first and second sounds = 14 minutes 30 seconds.
• Speed of sound v_s = 330 m/s.
• The bus moves directly toward the firing point at constant speed.
• Sound travels in straight line from the gun to the bus.
• We ignore wind and other effects.

Concept / Approach:
Let d1 and d2 be the distances between the bus and the gun at the instants of the first and second shots. The times for sound to reach Rahul are t1 = d1 / v_s and t2 = d2 / v_s. We know the difference between the firing interval and the hearing interval, so we can relate d1 and d2. The bus moves closer, so d1 - d2 equals bus speed multiplied by 15 minutes of travel. Solving these equations yields the bus speed in m/s.

Step-by-Step Solution:
Let the bus speed be v_b m/s and sound speed be v_s = 330 m/s. Time between shots: 15 minutes = 15 * 60 = 900 seconds. Time between hearing the sounds: 14 minutes 30 seconds = 870 seconds. Let the distances at the instants of firing be d1 and d2 metres. Then hearing times are: first shot at t = d1 / 330, second shot at t = 900 + d2 / 330. Given that the heard interval is 870 seconds, we have (900 + d2 / 330) - (d1 / 330) = 870. Simplify: 900 - 870 = (d1 - d2) / 330, so 30 = (d1 - d2) / 330, hence d1 - d2 = 30 * 330 = 9900 metres. The bus moves toward the gun during the 900 seconds between shots, so d1 - d2 = v_b * 900. Set v_b * 900 = 9900 and solve for v_b: v_b = 9900 / 900 = 11 m/s.

Verification / Alternative check:
Convert the bus speed to km/h: 11 m/s * 3.6 = 39.6 km/h. Using this speed, the bus travels 39.6 * (1/4) = 9.9 km, which is 9900 m, in 15 minutes. This matches the calculated difference d1 - d2, consistent with the sound timing difference of 30 seconds between the actual firing interval and the heard interval.

Why Other Options Are Wrong:
Speeds of 12, 10, 9 or 8 m/s would make d1 - d2 equal to 10800, 9000, 8100 or 7200 metres respectively, which would not match the relation d1 - d2 = 9900 metres implied by the sound timing. Therefore these options contradict the timing information.

Common Pitfalls:
A frequent error is to use 15 minutes instead of the difference between 15 minutes and 14 minutes 30 seconds when relating distances. Another mistake is failing to convert all times to seconds before using the speed of sound in m/s. Keeping units consistent and clearly distinguishing between firing times and hearing times is essential for solving such problems.

Final Answer:
The approximate speed of the bus is 11 m/s.