Fundamentals of Physics Extended (10th Edition)

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

Chapter 2 - Motion Along a Straight Line - Problems - Page 34: 34a

Answer

$v_0 = -47.9~km/h$

Work Step by Step

We can convert $20~km/h$ to units of $m/s$: $(20~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 5.56~m/s$ We can find the time it takes the red car to reach $x = 44.5~m$: $t = \frac{44.5~m}{5.56~m/s} = 8.0~s$ We can convert $40~km/h$ to units of $m/s$: $(40~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 11.11~m/s$ We can find the time it takes the red car to reach $x = 76.6~m$: $t = \frac{76.6~m}{11.11~m/s} = 6.9~s$ We can write an equation for the green car at $t = 8.0~s$: $x-x_0 = v_0~t+\frac{1}{2}at^2$ $44.5-220 = 8.0~v_0+\frac{1}{2}a(8.0)^2$ $8.0~v_0+32~a+175.5 = 0$ We can write an equation for the green car at $t = 6.9~s$: $x-x_0 = v_0~t+\frac{1}{2}at^2$ $76.6-220 = 6.9~v_0+\frac{1}{2}a(6.9)^2$ $6.9~v_0+23.8~a+143.4 = 0$ $(-\frac{32}{23.8})\times(6.9~v_0+23.8~a+143.4) = 0$ $-9.3~v_0-32~a-192.8 = 0$ We can add the two equations for the green car: $-1.3~v_0-17.3 = 0$ $v_0 = -\frac{17.3}{1.3}$ $v_0 = -13.3~m/s$ $v_0 = (-13.3~m/s)(\frac{3600~s}{1~h})(\frac{1~km}{1000~m})$ $v_0 = -47.9~km/h$
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