Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 51: 48


(a) $77.8~m$ (b) The cheetah's speed is 12.5 m/s. (c) $6.25~m/s$ (d) The first 3.11 s: $19.4~m$ The second 3.11 s: $58.4~m$

Work Step by Step

(a) $a_{av}=\frac{1}{2}(v_f+v_0)=\frac{1}{2}(25.0~m/s+0)=12.5~m/s$ $x=x_0+v_{av}t$. Now, assuming that the starting position is at the origin ($x_0=0$): $x=0+(12.5~m/s)(6.22~s)=77.8~m$ (b) The velocity, in a motion with constant acceleration, varies linearly with time. That is, starting from zero, the velocity is directly proportional to the time. But, $3.11~s=\frac{6.22~s}{2}$. Now, we can conclude that, after sprinting for 3.11 s, the cheetah's speed is $12.5 m/s=\frac{25.0~m/s}{2}$. (c) $a_{av}=\frac{1}{2}(v_f+v_0)=\frac{1}{2}(12.5~m/s+0)=12.5~m/s=6.25~m/s$ (d) The first 3.11 s: $x=x_0+v_{av}t$ $x=0+(6.25~m/s)(3.11~s)=19.4~m$ The second 3.11 s: In the 6.22 the cheetah has run 77.8 m. So in the second 3.11 s: $\Delta x=77.8~m-19.4~m=58.4~m$
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