Answer
(a) The magnitude of the acceleration is $|a|=5.3~m/s^2$ and in the opposite direction of the chicken's movement.
(b) $3.2~m$
Work Step by Step
(a) $v_0=5.8~m/$, $v=0$, $\Delta t=1.1~s$
$v=v_0+at$
$0=5.8~m/s+a(1.1~s)$
$-(1.1~s)a=5.8~m/s$
$a=\frac{5.8~m/s}{-1.1~s}=-5.3~m/s^2$.
$a$ and $v_0$ have opposite signs.
The magnitude of the acceleration is $|a|=5.3~m/s^2$ and in the opposite direction of the chicken's movement.
(b) $v_{av}=\frac{1}{2}(v+v_0)=\frac{1}{2}(0+5.8~m/s)=2.9~m/s$
$v_{av}=\frac{\Delta x}{\Delta t}$. Rearrange the equation:
$\Delta x=(v_{av})(\Delta t)=(2.9~m/s)(1.1~s)=3.2~m$
