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 49: 24

Answer

(a) See the graph. (b) $v_{av}=-2~m/s$ (c) $average~speed=2.16~m/s$
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Work Step by Step

(a) See the graph. (b) $x=(-5~m/s)t+(3~m/s^{2})t^{2}$ For $t=0$: $x=(-5~m/s)0+(3~m/s^{2})0^{2}=0$ For $t=1~s$: $x=(-5~m/s)(1~s)+(3~m/s^{2})(1~s)^{2}=-5~m+3~m=-2~m$ $v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{-2~m-0}{1~s-0}=-2~m/s$ (c) From mathematics: The equation of a parabola: $y=ax^2+bx+c$ The vertex: (h, k), where $h=-\frac{b}{2a}$ and $k=\frac{4ac-b^{2}}{4a}$ (Let $t_v$ be the time in the vertex of the parabola and $x_v$ the position in the vertex of the parabola). To physics: ($y~-> x,~x~->t,~h~->t_v,~k~->x_v$) The equation of a parabola : $x=at^2+bt+c=c+bt+at^2$ The vertex: ($t_v,~x_v$), where $t_v=-\frac{b}{2a}$ and $x_v=\frac{4ac-b^{2}}{4a}$ Notice that the $a$ in the parabola equation does not means acceleration. Let's find the vertex of the parabola, that is, let's find the moment ($t_v$) when the particle reaches the lowest position ($x_v$). $t_v=-\frac{b}{2a}=-\frac{-5~m/s}{2\times3~m/s^{2}}=0.83~s$. $t_v$ is between $t=0$ and $t=1~s$ $4ac-b^{2}=4(-2~m/s^{2})0-(-5~m/s)^{2}=-25~m^{2}/s^{2}$ $x_v=\frac{4ac-b^{2}}{4a}=\frac{-25~m^{2}/s^{2}}{4\times3~m/s^{2}}=-2.08~m$ The particle moves 2.08 m in the negative direction (from $t=0$ to $t=0.83~s$). Then it moves 0.08 m in the positive direction (from $t=0.83~s$ to $t=1~s$). So, the total distance traveled is equal to 2.16 m. Now, let's find the average speed: $average~speed=\frac{distance}{elapsed~time}=\frac{2.16~m}{1~s}=2.16~m/s$
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