Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 52: 63

Answer

(a) $0.25~s$ (b) $110~m/s^2$ (c) Heigth: $0.55~m$ Speed: $11~m/s$

Work Step by Step

(a) $v_{av}=\frac{v+v_0}{2}=\frac{26.0~m/s+0}{2}=13.0~m/s$ $v_{av}=\frac{\Delta x}{\Delta t}$. Rearrange the equation: $\Delta t=\frac{\Delta x}{v_{av}}=\frac{3.2~m}{13.0~m/s}=0.25~s$ (b) $v^2=v_0^2+2a\Delta x$ $(26.0~m/s)^2=0^2+2a(3.2~m)$ $a=\frac{676~m^2/s^2}{6.4~m}=110~m/s^2$ (c) Put the origin at the starting position: $x_0=0$ $x=x_0+v_0t+\frac{1}{2}at^2$. $x=0+0(0.10~s)+\frac{1}{2}(110~m/s^2)(0.10~s)^2$ $x=0.55~m$ $v=v_0+at=0+(110~m/s^2)(0.10~s)=11~m/s$
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