Answer
a) The body's displacement is $-20m$ and its average velocity is $-5m/s$.
b) At $t=-4$: speed $=25m/s$ and acceleration $=50m/s^2$.
At $t=0$: speed $=1m/s$ and acceleration $=0.4m/s^2$.
c) The body does not change direction in the given interval.
Work Step by Step
$$s(t)=\frac{25}{t+5}$$
a) The body's displacement, $\Delta s$, over the interval $[-4,0]$ is $$\Delta s=s(0)-s(-4)=\frac{25}{0+5}-\frac{25}{-4+5}=5-25=-20m$$
The average velocity, $v_a$, of the body over the interval $[-4,0]$ is $$v_a=\frac{\Delta s}{0-(-4)}=\frac{-20}{4}=-5m/s$$
b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}\Big(\frac{25}{t+5}\Big)=\frac{-25(t+5)'}{(t+5)^2}=-\frac{25}{(t+5)^2}$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}\Big(-\frac{25}{(t+5)^2}\Big)=-\frac{-25\Big((t+5)^2\Big)'}{(t+5)^4}$$ $$a(t)=\frac{25(t^2+10t+25)'}{(t+5)^4}=\frac{25(2t+10)}{(t+5)^4}=\frac{50t+250}{(t+5)^4}$$
We now examine the speed and acceleration at 2 endpoints of $[-4,0]$:
- At $t=-4$: $$v(-4)=-\frac{25}{(-4+5)^2}=-25m/s$$
So the speed of the body is $|v(-4)|=25m/s$
$$a(-4)=\frac{50(-4)+250}{(-4+5)^4}=50m/s^2$$
- At $t=0$: $$v(0)=-\frac{25}{(0+5)^2}=-1m/s$$
So the speed of the body is $|v(0)|=1m/s$
$$a(0)=\frac{50\times0+250}{(0+5)^4}=0.4m/s^2$$
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 $[-4,0]$ 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 $$v(t)=\frac{-25}{(t+5)^2}$$
This function, however, never reaches $0$ no matter which $t$ there is. It is also always negative.
Therefore, $v(t)$ does not ever change signs, meaning that the body does not ever change direction also.
