Physics (10th Edition)

Published by Wiley

Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 42

Answer

a) 13.1 m/s b) 0.94 $m/s^{2}$

Work Step by Step

Let the car's velocity = b $distance = speed \times time$ At 14s, $d=(b)(14)=14b$ At 28s, $d=(b)(28)=28b$ Let train's acceleration = a $x=ut+\frac{1}{2}at^{2}$ At 14s, $x=0+\frac{1}{2}a(14)^{2}=98a$ At 28s, $x=0+\frac{1}{2}a(28)^{2}=392a$ At 14s, the car has traveled the same distance as the train + length of train $14b=98a-92.......equation\;1$ At 28s, the car has traveled the same distance as the train $28b=392a...........equation\; 2$ We now use a system of linear equations: $28b=196a-184........equation\; 1 \times$2 Sub in equation 2: $(392a)=196a-184$ $196a=184$ $a=\frac{46}{49}\approx0.94\; m/s^{2}$ $28b=392a$ $28b=392(\frac{46}{49})$ $b=\frac{92}{7}\approx13.1$ m/s

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