Answer to Question #138241 in Physics for N/a

Question

A cheetah is walking at 1.0m/s when it sees a  zebra 25m away. What acceleration would be  required to reach 20.0m/s in that distance? 

Answer 1

The required acceleration is

$$a=\frac{v_f^2-v_i^2}{2d}=\frac{20^2-1^2}{2\cdot25}=7.98\text{ m/s}^2.$$