Book edition 9th Edition

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

Pages 1624 pages

ISBN 9781133947271

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Equation of freely falling body

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

Step 2: Explanation for correct answer

In first case, (First 1s)

Travels 4.00 m in the first 1s.

$4.00 m = \frac{1}{2} a_{g} (1 s)^{2} a_{g} = 8.00 m / s^{2}$

In second case, (after 2s )

$x_{f} = \frac{1}{2} (8.00 m / s^{2}) (2 s)^{2} = 16 m$

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

$16.00 m - 4.00 m = 12 m$ .

Hence, the answer is 12 m.

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Physics For Scientists & Engineers

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Equation of freely falling body

Step 2: Explanation for correct answer

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Physics For Scientists & Engineers

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Equation of freely falling body

Step 2: Explanation for correct answer

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