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Expert-verified Found in: Page 49 ### Physics For Scientists & Engineers

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 1s ? (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.

See the step by step solution

## Step 1: Equation of freely falling body

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

## Step 2: Explanation for correct answer

In first case, (First 1s)

Travels 4.00 m in the first 1s.

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

In second case, (after 2s )

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

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

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

Hence, the answer is 12 m. ### Want to see more solutions like these? 