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: 38a

Answer

$32.9m/s$

Work Step by Step

To reach the maximum speed, the subway can accelerate for half of the distance and decelerate for the second half. This means that the distance over the acceleration is $\frac{806m}{2}=403m$. A formula relating distance, acceleration, initial velocity and displacement is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $v_f$ yields $$v_f=\sqrt{v_o^2+2a\Delta x}$$ Using values of $v_o=0.00m/s$, $a=1.34m/s^2$, and $\Delta x=403m$, the final velocity is equal to $$v_f=\sqrt{(0.00m/s)^2+2(1.34m/s^2)(403m)}$$ $$v_f=32.9m/s$$
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