College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 63: 51

Answer

(a) See step by step solution. (b) 96 (m)

Work Step by Step

(a)Let's assume for simplicity that initial velocity is zero ($v_{0}=0$) and the body starts moving from a point $x_{0}=0$. Formulas for motion with constant acceleration are $$v=at,$$ $$s=\dfrac{at^2}{2}.$$ The graph $v(t)$ is shown in the Figure 1. The shaded area (the triangle) is equal to $S=\dfrac{1}{2}vt$. The velocity at time t is equal to $v=at$. So $S=\dfrac{1}{2}at^2.$ So the area is equal to the displacement. (b) We need to calculate the distance traveled by the body, the graph of which is shown in Figure 2. The distance (as we know from (a)) is equal to the area under the curve of a velocity-versus-time plot. So we have to find that area. The area of ​​this figure is equal to the area of ​​the trapezoid. The area of the trapezoid is $S=\dfrac{a+b}{2}h$ where $a$ and $b$ are the lengths of the parallel sides, $h$ is the height. In our case $a=10.0 (s)-4.0(s)=6.0(s)$; $b$ is the total travel time. $b=18(s)$; $h$ is the highest velocity value. $h=8 (m/s)$. So the area or the displacement is equal to $$S=\dfrac{6(s)+18(s)}{2}8(m/s)=96 (m)$$
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