Question

A man runs at a speed of 4 m/s to overtake a standing bus. When he is 6 m behind the door at \( t = 0 \), the bus moves forward and continuous with a constant acceleration of 1.2 m/s². The man reaches the door in time t. Then,

• A. \( 4t = 6 - 0.6t^2 \)
• B. \( 4t = 1.2t^2 \)
• C. \( 4t = 1.2t + 4t^2 \)
• D. \( 4t = 6 + 4t^2 \)

Hint:

In relative motion problems, ensure you account for the motion of both objects and set their positions equal at the time of interest.

Answer 1

The Correct Option is A

Step 1: Understand the relative motion between the man and the bus.
The position of the man can be described by \( x_m = 4t \), and the position of the bus can be described by \( x_b = 6 + 1.2t^2 \).
Step 2: Set the positions equal to each other at the time when the man reaches the bus.
We equate the distances: \[ 4t = 6 + 1.2t^2 \]
Step 3: Rearrange the equation.
\[ 4t = 6 - 0.6t^2 \] Final Answer: \[ \boxed{4t = 6 - 0.6t^2} \]