Answer
a. $-20m$ and$-5m/s$.
b. $25m/s$ and $1m/s$.
$50m/s^2$ and $\frac{2}{5}m/s^2$.
c. no change in direction.
Work Step by Step
a. Given $s(t)=\frac{25}{t+5}$ and $-4\leq t\leq0$, we have $\Delta t=4s$, $s(-4)=25m$ and $s(0)=5m$. Thus the body’s displacement is $\Delta s=s(0)-s(-4)=-20m$ and the average velocity for the given time interval is $\bar v=\Delta s/\Delta t=-20/4=-5m/s$.
b. Velocity $v(t)=s'(t)=\frac{-25}{(t+5)^2}$; thus the speeds at the endpoints are $|v(-4)|=|\frac{-25}{(-4+5)^2}|=25m/s$ and $|v(0)|=|\frac{-25}{(0+5)^2}|=1m/s$.
Acceleration is $a(t)=v'(t)=\frac{50}{(t+5)^3}$; thus $a(-4)=50m/s^2$ and $a(0)=50/5^3=\frac{2}{5}m/s^2$.
c. When the body changes its direction, its velocity will pass $0$ in either direction. Letting $v(t)=0$, we have $\frac{-25}{(t+5)^2}=0$; as there is no solution to this equation, the body will not change its direction.
