Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 157: 56

Answer

$a= 26.15 ms^{-2}$

Work Step by Step

let $a$ be the acceleration of the rocket for $16s$ and $S_1$ be the distance(height) travelled in $16s$. From $16s$ to $20s$ acceleration of the rocket is $-g$ and $S_2$ be the distance travelled in $20s -16s =4s$. Let $v$ be the velocity after $16s$ Now $v = u +at = 0 + 16a $ $v =16a$ $S_1 =ut +\frac{1}{2}at^2$ $ S_1 = 0 + 128a $ $S_1= 128a$ $S_2 =vt -\frac{1}{2}gt^2$ $S_2 = 4v- 80$ we have $S_1 + S_2 =5100m$ , therefore $5100m = 128 a + 4 v -80$ putting value of $v$ $5100 m = 128a +64a -80$ $a= 26.15 ms^{-2}$
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