Answer to Question #133144 in General Chemistry for Angelo Cimmarrusti

Question

A hawk is flying in the sky and wants to dive to catch a mouse running through a field. The hawk is initially moving at 1.25 m/s but as it dives to catch the mouse, it speeds up to 13.50 m/s in only 2.00 seconds. What is the acceleration of the hawk?

Answer 1

Acceleration of the hawk:

$a=(V_2-V_1)/\Delta t=(13.50m/s-1.25m/s)/2.00s=$

$=12.25m/s/2.00s=6.125m/s^2$