A starts from a point with a speed of 30 metres/second. After 3 seconds, B starts chasing A from the same point with a speed of 50 metres/second. What will be the total distance (in metres) travelled by A and B together before A is caught by B?

Introduction / Context:
This is a relative speed and catch-up problem in metres and seconds. One runner, A, starts first, and after a delay B starts from the same point with a higher speed. Because B is faster, he will eventually catch up with A. The question asks for the total distance travelled by both runners combined up to the moment of catching, not just the distance from the starting point to the meeting point. This twist requires careful thinking after solving for the catch-up time.

Given Data / Assumptions:
- Speed of A = 30 m/s.
- Speed of B = 50 m/s.
- A starts at t = 0 seconds from a point.
- B starts 3 seconds later from the same point, at t = 3 seconds.
- Both run along the same straight line in the same direction with constant speeds.
- We need the sum of the distances A and B have individually travelled at the instant B catches A.

Concept / Approach:
First, we find the time at which B catches A. Let t be the time (in seconds) from A’s start until they meet. During this time, A runs for t seconds and B runs for (t - 3) seconds. They are at the same position at the catch-up moment, so their distances from the starting point are equal. This gives an equation in t. Once we solve for t, we can compute the distance each has travelled from the start and then add those two distances to get the total distance travelled by both runners together.

Step-by-Step Solution:
Step 1: Let t be the time in seconds from A’s start until B catches A.Distance covered by A in time t: d_A = 30 * t metres.Distance covered by B in time t - 3: d_B = 50 * (t - 3) metres.Step 2: At the catch-up moment, d_A = d_B.So 30 * t = 50 * (t - 3).Step 3: Solve for t: 30t = 50t - 150 ⇒ 20t = 150 ⇒ t = 150 / 20 = 7.5 seconds.Step 4: Compute the distance from the start to the catch point.d = 30 * 7.5 = 225 metres (A’s distance; B’s will match).Step 5: Total distance travelled by both runners together is d_A + d_B = 225 + 225 = 450 metres.

Verification / Alternative check:
Check B’s distance using t = 7.5 seconds. B’s running time is t - 3 = 4.5 seconds, so d_B = 50 * 4.5 = 225 metres. This matches A’s distance, confirming the catch-up point. The total combined distance of 450 metres is therefore just twice 225 metres, consistent with both having reached the same spot at the same time.

Why Other Options Are Wrong:
- 360 metres, 600 metres, and 720 metres do not correspond to 2 * 225 metres. To reach these totals, the runners would have to meet at distances different from 225 metres or at different times, which would contradict the equality of path lengths at the catch-up moment.

Common Pitfalls:
Many students mistakenly give 225 metres as the answer, thinking only about the distance from the origin to the meeting point. However, the question explicitly asks for the total distance travelled by both A and B, so we must add the two individual distances. Another common error is forgetting that B runs for (t - 3) seconds, not t seconds, or mixing up speeds and times when solving the equation.

Final Answer:
The total distance travelled by A and B before A is caught is 450 metres.