Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 2 - Kinematics in One Dimension - Problems - Page 52: 79

Answer

$$a =6 \mathrm{m} \cdot \mathrm{s}^{-2}$$

Work Step by Step

To find the acceleration of the cheetah, we need to find the distance that the cheetah would travel and the distance from its prey, and then we must equate them to each other. Thus: $$ v_{p} t=x $$ where $v_{c}$ is the velocity of the prey. The cheetah will start from rest (which means $0=v_{0}$ ) with an accelration of $a,$ thus the distance that the cheetah moves is: $$ \begin{array}{c} v_{0} t+\frac{1}{2} a t^{2}=x \\ \frac{1}{2} a t^{2}=x \end{array} $$ by equating (1) and (2), we get: $$ \frac{1}{2} a t^{2}=v_{p} t $$ $$ \frac{2 v_{p}}{t}=a $$ Plugging in given values: $$ \begin{aligned} &\frac{2\left(9 \mathrm{m} \cdot \mathrm{s}^{-1}\right)}{(3 \mathrm{s})}=a \\ &=6 \mathrm{m} \cdot \mathrm{s}^{-1} \\ 6 \mathrm{m} \cdot \mathrm{s}^{-2}=a & \end{aligned} $$
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