Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems: 30

Answer

(a) 440 meters (b) 42 seconds (c) first second: 21 meters fifth second: 19 meters

Work Step by Step

(a) Let's convert 75 km/h to units of m/s. $75 ~km/h ~\times ~ 1000 ~ m/km ~\times ~1 h/3600 s = 21 ~m/s$ To find the distance the car coasts, we can use the equation $v^2= v_0^2 + 2a(x-x_0)$ $x = \frac{v^2 - v_0^2}{2a} = \frac{(-21)^2}{2(-0.50)}= 440 ~m$ (b) To find the time it takes to stop, we use the equation $v - v_0 = at$ Then $t = \frac{v-v_0}{a} = \frac{-21 ~m/s}{-0.50 ~m/s^2} = 42 ~s$ The time it takes to stop is 42 seconds. (c) The first second is from $t_1 = 0$ to $t_2 = 1$. $v_1 = 21 ~m/s$ and $v_2 = 21 - (0.50)(1) = 20.5 ~m/s$ The average velocity is $\frac{21+20.5}{2} = 20.75 ~m/s$ $x = 20.75 ~m/s \times 1 s = 21 ~m$ The distance it travels in the first second is 21 meters. The fifth second is from $t_4 = 4$ to $t_5 = 5$. $v_4 = 21 - (0.50)(4) = 19 ~m/s$ and $v_5 = 21 - (0.50)(5) = 18.5 ~m/s$ The average velocity is $\frac{19+18.5}{2} = 18.75 ~m/s$ $x = 18.75 ~m/s \times 1 s = 19 ~m$ The distance it travels in the fifthsecond is 19 meters. Note that the answers are rounded off to two significant digits.
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