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Q10OQ

Expert-verifiedFound in: Page 49

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**On another planet, a marble is released from rest at the top of a high cliff. It falls ** **4.00**** m in the first 1** **s of its motion. Through what additional distance does it fall in the next ** 1**s ? (a) ** **4.00**** m (b)**** 8.00 ****m (c) ****12.0** **m (d) **** 16.0** **m (e) 20.0** **m**

So, the correct answer is option (c) 12.0 m.

**Vertical displacement of a freely falling body is given by the relation,** ${\mathit{x}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathit{a}}}_{{\mathbf{g}}}{{\mathit{t}}}^{{\mathbf{2}}}$ , where Planet ${a}_{g}=$acceleration.

In first case, (First 1s)

Travels 4.00 m in the first 1s.

$4.00\text{m}=\frac{1}{2}{a}_{g}(1\text{s}{)}^{2}\phantom{\rule{0ex}{0ex}}{a}_{g}=8.00m/{s}^{2}\phantom{\rule{0ex}{0ex}}$

In second case, (after 2s )

${x}_{f}=\frac{1}{2}(8.00m/{s}^{2})(2\text{s}{)}^{2}\phantom{\rule{0ex}{0ex}}=16\text{m}$

Hence, after 2 seconds of travel from rest, the change in position is 16.00 m.

So, the change in position between 1s and 2s is

$16.00\u200am-4.00\text{m}=12\u200am$.

Hence, the answer is 12 m.

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