Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 85: 18a

Answer

$\left|v\right|=15.8m/s$

Work Step by Step

$a_{x}=5.00m/s^{2}$ is constant, so we can use Table 2-1 in the x-direction. $a_{y}=7.00m/s^{2}$ is constant, so we can use Table 2-1 in the y-direction.. $\left[\begin{array}{ll} v=v_{0}+at, & (2- 11)\\ x-x_{0}=v_{0}t+\frac{1}{2}at^{2}, & (2- 15) \\ v^{2}=v_{0}^{2}+2a(x-x_{0}) , & (2- 16) \\ x-x_{0}=\frac{1}{2}(v_{0}+v)t, & (2- 17) \\ x-x_{0}=vt-\frac{1}{2}at^{2} & (2- 18) \end{array}\right]$ Given $x-x_{0}=12.0m, \quad v_{x0}=4.0m/s,\quad a_{x}=5.00m/s^{2},$ we solve $(2- 15)$ for $t:$ $12.0 m=(4.00m/s)t+\displaystyle \frac{1}{2}(5.00m/\mathrm{s}^{2})t^{2}$ $(2.50m/\mathrm{s}^{2})t^{2}+(4.00m/s)t-12.0 m=0$ The quadratic formula gives $t=\displaystyle \frac{(4.00m/s)\pm\sqrt{(4.00m/s)-4(-12.0 m)(2.50m/\mathrm{s}^{2})}}{2(2.50m/\mathrm{s}^{2})}$ discarding the negative solution, gives $t=1.53\mathrm{s}$. We now find components of velocity at $t=1.53\mathrm{s}$. In the x direction , Eq. 2-11 gives $v_{x}=v_{x0}+a_{x}t=4.00m/s+(5.00m/s^{2})(1.53\mathrm{s})=11.7m/s$ In the $y$ direction , Eq. 2-11 gives $v_{y}=v_{y0}+a_{y}t=0+(7.00m/s^{2})(1.53\mathrm{s})=10.7m/s$ Thus, $\qquad \vec{v}=(11.7m/s)\hat{i}+(10.7m/s)\hat{j}$ (a) Its magnitude is $\left|v\right|=\sqrt{(11.7m/s)+(10.7m/s)}=15.8m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.