University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 59: 2.15


a) $x_{0}=50 cm$ $v_{0} = 2 cm/s$ $a_{0}=-0.125cm/s^{2}$ b) $t=16s$ c) $t=32s$ after starting d) $t\approx 6.2s \Longrightarrow v(t\approx6.2s)\approx 1.22cm/s$ $t\approx 25.8s \Longrightarrow v(t\approx 25.8s)\approx -1.22cm/s$ $t\approx -4.4s \Longrightarrow v(t\approx -4.4s)\approx 2.55 cm/s$ (see the work step by step for more details) $t\approx 36.4s \Longrightarrow v(t\approx 36.4s)\approx -2.55 cm/s$

Work Step by Step

\approxa) The equation for the turtle's position as a function of time is a quadratic equation. That means that the acceleration is constant. When the acceleration is constant, $x(t)$ is given by $x(t)=x_{0}+v_{0}t-\frac{a}{2}t^{2}$. We can make links between the given equation and this general form : $x_{0}=50cm$ $v_{0}=2 cm/s$ $\frac{a}{2}=-0.0625 \Longleftrightarrow a=-0.125cm/s^{2}$ b) We find the time $t$ when $v(t)=0$. In such a motion, $v(t)=v_{0}+at$. We can simply solve the equation : $v(t)=2-0.125t=0$ $t=16s$ c)We find the time $t$ when $x(t)=x_{0}=50cm$. We can simply solve the equation : $x(t)=50+2t-0.0625t^{2}=50$ $-0.0625t^{2}+2t=0$ $t(-0.0625t+2)=0$ We have 2 possibilities : $t=0s$ or $-0.0625t+2=0$. The question asks when the turtle $returns$ to its starting point. So $t>0$ and we eliminate the first option. The second option gives : $t=32s$. d) The turtle is a distance of 10.0 cm from its starting point when $x=40cm$ and when $x=60cm$. Let's first determine at which times $x=40cm$ : $-0.0625t^{2}+2t+50=40$ $-0.0625t^{2}+2t+10=0$ $\Delta=6.5$ $t_{1}\approx-4.4s$ ; $t_{2}\approx 36.4s$ IF the given equation is defined for $t<0$, we can accept $t_{1}$ as an answer. The domain of the equation is not specified, but a negative time doesn't really makes sense in this case... Let's now determine when $x=60cm$ : $-0.0625t^{2}+2t+50=60$ $-0.0625t^{2}+2t-10=0$ $\Delta=1.5$ $t_{3}\approx6.2s$ ; $t_{4}\approx 25.8s$ We still have to calculate the velocities at each of those times... $v(t)=v_{0}+at=2-0.125t$ $v(t_{1})\approx2.55cm/s$ $v(t_{2})\approx -2.55cm/s$ $v(t_{3})\approx 1.22cm/s$ $v(t_{4})\approx -1.22cm/s$
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