$ v_{\text{escalator}} $ as the speed of the escalator.

When the girl walks up on the stationary escalator, she takes 20 seconds. This gives us:

$$ d = v_{\text{girl}} \times 20 \, \text{s} $$

When the escalator alone moves and the girl does not walk, the time taken is 30 seconds. Thus:

$$ d = v_{\text{escalator}} \times 30 \, \text{s} $$

When the girl walks up on the moving escalator, both her speed and the escalator's speed contribute to her total speed:

$$ d = (v_{\text{girl}} + v_{\text{escalator}}) \times t $$

Setting the distances equal since they all cover distance $ d $:

From $ v_{\text{girl}} $:

$$ v_{\text{girl}} = \frac{d}{20} $$

From $ v_{\text{escalator}} $:

$$ v_{\text{escalator}} = \frac{d}{30} $$

Combining both in the case of the moving escalator:

$$ d = \left(\frac{d}{20} + \frac{d}{30}\right) \times t $$

Simplifying the combination:

$$ 1 = \left(\frac{1}{20} + \frac{1}{30}\right) \times t $$

Finding a common denominator and solving:

$$ \frac{1}{20} + \frac{1}{30} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} $$

Now, solve for $ t $:

$$ 1 = \frac{1}{12} \times t $$

So:

$$ t = 12 \, \text{s} $$

Hence, the time taken by her to walk up on the moving escalator is 12 seconds.

Option C: 12 s is the correct answer.

Motion In A Straight Line

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