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### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 s? (a) 9.8 m (b) 19.6 m (c) 39 m (d) 44 m (e) none of the above.

Hence, option (c) 39 m is correct.

See the step by step solution

## Step 1: Definition of speed

The magnitude of an object's rate of change of position with time, or the magnitude of change of position per unit of time, is its speed. It is thus a scalar quantity.

## Step 2: Explanation for correct answer

Given:

Initial velocity ${v}_{0}=0$

The distance travelled in 1s is $\Delta {x}_{1}=4.9$ m

Let $x$ be the distance travelled in the next 2s

Total distance travelled in $3 \text{s}=4.9 \text{m}+x$

From the equation of motion, $\Delta x={v}_{e}t+\frac{1}{2}a{t}^{2}$

role="math" localid="1663684119423" $4.9\text{m}+x=\left(\frac{1}{2}\right)\left(9.8m/{s}^{2}\right)\left(3\text{s}{\right)}^{2}\phantom{\rule{0ex}{0ex}}x=\left(\frac{1}{2}\right)\left(9.8m/{s}^{2}\right)\left(3\text{s}{\right)}^{2}\phantom{\rule{0ex}{0ex}}x=39.2\text{m}\phantom{\rule{0ex}{0ex}}\approx 39\text{m}$

Therefore, at approximately 39 m the pebble dropped in the next 2.0 s.

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