is relative speed results in an increase in the distance between them by $$6 \mathrm{~m}$$ in one second. Therefore, we have:

$$ |v_1 - v_2| = 6 $$

When the two objects are moving towards each other, the relative speed is given by:

$$ v_1 + v_2 $$

In this case, the distance decreases by $$8 \mathrm{~m}$$ in one second, so we have:

$$ v_1 + v_2 = 8 $$

We now have two equations:

We can solve these equations by considering the absolute value condition in two cases:

Case 1: $$v_1 \ge v_2$$

$$ v_1 - v_2 = 6 \quad \text{(from |v_1 - v_2| = 6 )} $$

$$ v_1 + v_2 = 8 $$

Adding these two equations:

$$ (v_1 - v_2) + (v_1 + v_2) = 6 + 8 $$

$$ 2v_1 = 14 $$

$$ v_1 = 7 $$

Substituting $$v_1 = 7$$ into $$v_1 + v_2 = 8$$:

$$ 7 + v_2 = 8 $$

$$ v_2 = 1 $$

Case 2: $$v_1 < v_2$$

$$ v_2 - v_1 = 6 \quad \text{(from |v_1 - v_2| = 6 )} $$

$$ v_1 + v_2 = 8 $$

Adding these two equations:

$$ (v_2 - v_1) + (v_1 + v_2) = 6 + 8 $$

$$ 2v_2 = 14 $$

$$ v_2 = 7 $$

Substituting $$v_2 = 7$$ into $$v_1 + v_2 = 8$$:

$$ v_1 + 7 = 8 $$

$$ v_1 = 1 $$

From both cases, we find that the speeds of the objects are $$7 \mathrm{~ms}^{-1}$$ and $$1 \mathrm{~ms}^{-1}$$.

Therefore, the correct answer is:

Option B: $$7 \mathrm{~ms}^{-1}$$ and $$1 \mathrm{~ms}^{-1}$$

Motion

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