## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v=\sqrt{9.17s-27.7}$ ft/s
We solve for the acceleration as follows: $v^2=v_0^2+2a_c(s-s_0)$ $8^2=3^2+2a_c(10-4)$ $a_c=4.583$ ft/s$^2$ Next, we solve for velocity: $v^2=v_0^2+2a_c(s-s_0)$ $v^2=3^2+2(4.583)(s-4)$ $v=\sqrt{9.17s-27.7}$ ft/s