University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 62: 2.56


(a) The initial velocity is -60.0 m/s (b) v = -15.5 m/s

Work Step by Step

Let up be the positive direction. $y(t) = b - ct + dt^2$ $y(t) = (800)~m - (60.0~m/s)~t + (1.05~m/s^2)~t^2$ (a) $v(t) = \frac{dy}{dt} = (-60.0~m/s)+ (2.10~m/s^2)~t$ When t = 0 s, $v = (-60.0~m/s)+ (2.10~m/s^2)(0) = -60.0~m/s$ The initial velocity of the lander is -60.0 m/s (b) Let $v$ be the velocity just before landing. $v^2 = v_0^2+2a(y-y_0)$ $v = \sqrt{v_0^2+2a(y-y_0)}$ $v = \sqrt{(-60.0~m/s)^2+(2)(2.10~m/s^2)(-800~m)}$ $v = -15.5~m/s$ Note that the negative sign means that the lander is descending onto the lunar surface.
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