Question

When two objects are moving along a straight line in the same direction, the distance between them increases by 6 m in one second. If the objects move with their constant speed towards each other, the distance decreases by 8 m in one second. Then the speed of the objects are:

• A. 14 m/s, 2 m/s
• B. 7 m/s, 1 m/s
• C. 3.5 m/s, 2 m/s
• D. 3.5 m/s, 1 m/s

Hint:

When solving relative speed problems, use the equations for relative speed in the same direction and when moving towards each other. Add or subtract the speeds as necessary to solve for the individual speeds.

Answer 1

The Correct Option is B

Let the speeds of the two objects be \( v_1 \) and \( v_2 \).
Step 1: When the objects are moving in the same direction
The relative speed of the two objects when moving in the same direction is \( |v_1 - v_2| \). According to the problem, the distance between the objects increases by 6 m in one second, so we have: \[ |v_1 - v_2| = 6 \, \text{m/s}. \]
Step 2: When the objects are moving towards each other
When the objects move towards each other, the relative speed is \( v_1 + v_2 \). In this case, the distance decreases by 8 m in one second, so we have: \[ v_1 + v_2 = 8 \, \text{m/s}. \]
Step 3: Solve the system of equations
We now have the following system of equations: 1. \( |v_1 - v_2| = 6 \) 2. \( v_1 + v_2 = 8 \) From equation (1), we have two cases: - Case 1: \( v_1 - v_2 = 6 \) - Case 2: \( v_2 - v_1 = 6 \) # Case 1: \( v_1 - v_2 = 6 \) Solving the system of equations: \[ v_1 - v_2 = 6 \] \[ v_1 + v_2 = 8 \] Adding these two equations: \[ 2v_1 = 14 \quad \Rightarrow \quad v_1 = 7 \, \text{m/s}. \] Substitute \( v_1 = 7 \) into \( v_1 + v_2 = 8 \): \[ 7 + v_2 = 8 \quad \Rightarrow \quad v_2 = 1 \, \text{m/s}. \] Thus, the speeds of the two objects are \( v_1 = 7 \, \text{m/s} \) and \( v_2 = 1 \, \text{m/s} \). Therefore, the correct answer is \( 7 \, \text{m/s} \) and \( 1 \, \text{m/s} \), which corresponds to option (B).