Answer
a. $-20m$ and$-5m/s$.
b. $45m/s$ and $\frac{1}{5}m/s$.
$140m/s^2$ and $\frac{4}{25}m/s^2$.
c. no change in direction over the interval.
Work Step by Step
a. Given $s(t)=\frac{25}{t^2}-\frac{5}{t}$ and $1\leq t\leq5$, we have $\Delta t=5-1=4s$, $s(1)=25-5=20m$ and $s(5)=\frac{25}{5^2}-\frac{5}{5}=0m$. Thus, the body’s displacement is $\Delta s=s(5)-s(1)=-20m$ and the average velocity for the given time interval is $\bar v=\Delta s/\Delta t=-20/4=-5m/s$.
b. Velocity $v(t)=s'(t)=-\frac{50}{t^3}+\frac{5}{t^2}$, thus speeds at the endpoints are $|v(1)|=|-\frac{50}{1^3}+\frac{5}{1^2}|=45m/s$ and $|v(5)|=|-\frac{50}{5^3}+\frac{5}{5^2}|=\frac{1}{5}m/s$.
Acceleration is $a(t)=v'(t)=\frac{150}{t^4}-\frac{10}{t^3}$; thus $a(1)=\frac{150}{1^4}-\frac{10}{1^3}=140m/s^2$ and $a(5)=\frac{150}{5^4}-\frac{10}{5^3}=\frac{4}{25}m/s^2$.
c. When the body changes its direction, its velocity will pass $0$ in either direction. Letting $v(t)=0$, we have $-\frac{50}{t^3}+\frac{5}{t^2}=0$ which gives $t=10s$ which is outside of the time interval. Thus the body will not change its direction in the interval.
