Answer
a) The body's displacement is $-20m$ and its average speed is $-5m/s$.
b) At $t=1$: speed $=-45m/s$ and acceleration $=140m/s^2$.
At $t=5$: speed $=-0.2m/s$ and acceleration $=0.16m/s^2$.
c) The body does not change direction in the given interval.
Work Step by Step
$$s(t)=\frac{25}{t^2}-\frac{5}{t}$$
a) The body's displacement, $\Delta s$, over the interval $[1,5]$ is $$\Delta s=s(5)-s(1)=(\frac{25}{5^2}-\frac{5}{5})-(\frac{25}{1^2}-\frac{5}{1})=(1-1)-(25-5)=-20m$$
The average velocity, $v_a$, of the body over the interval $[1,5]$ is $$v_a=\frac{\Delta s}{5-1}=\frac{-20}{4}=-5m/s$$
b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}\Big(\frac{25}{t^2}-\frac{5}{t}\Big)=\frac{-25(t^2)'}{t^4}-\Big(-\frac{5}{t^2}\Big)$$ $$v(t)=-\frac{50t}{t^4}+\frac{5}{t^2}=\frac{5}{t^2}-\frac{50}{t^3}$$
$$a(t)=\frac{dv}{dt}=\frac{d}{dt}\Big(\frac{5}{t^2}-\frac{50}{t^3}\Big)=\frac{-5(t^2)'}{t^4}-\Big(\frac{-50(t^3)'}{t^6}\Big)$$ $$a(t)=-\frac{10t}{t^4}+\frac{150t^2}{t^6}=\frac{150}{t^4}-\frac{10}{t^3}$$
We now examine the speed and acceleration at 2 endpoints of $[1,5]$:
- At $t=1$: $$v(1)=\frac{5}{1^2}-\frac{50}{1^3}=5-50=-45m/s$$
So the speed of the body is $|v(1)|=45m/s$
$$a(1)=\frac{150}{1^4}-\frac{10}{1^3}=150-10=140m/s^2$$
- At $t=5$: $$v(5)=\frac{5}{5^2}-\frac{50}{5^3}=\frac{1}{5}-\frac{2}{5}=-0.2m/s$$
So the speed of the body is $|v(5)|=0.2m/s$
$$a(5)=\frac{150}{5^4}-\frac{10}{5^3}=0.24-0.08=0.16m/s^2$$
c) We know that if
- $v(t)\gt0$: the body moves forward.
- $v(t)\lt0$: the body moves backward.
So the body will change direction when $v(t)$ changes sign. We will look for points in $[1,5]$ where $v(t)=0$, then examine the signs of $v(t)$ before and after that point to determine whether $v(t)$ changes sign there.
We have $$v(t)=\frac{5}{t^2}-\frac{50}{t^3}=\frac{5}{t^2}\Big(1-\frac{10}{t}\Big)$$
So $v(t)=0$ if $1-\frac{10}{t}=0$, or $t=10$.
$t=10$ does lie in the interval $[1,5]$, so we do not have to consider it.
Therefore, there is no point in $[1,5]$ for $v(t)$ to change signs, meaning that the body does not change direction in the given interval.