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Q12OQ

Expert-verifiedFound in: Page 49

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.

**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.**

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 $3\u200a\text{s}=4.9\u200a\text{m}+x$

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

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

Therefore, at approximately 39 m the pebble dropped in the next 2.0 s.94% of StudySmarter users get better grades.

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