Answer
The velocity of the particle at time $t=a$ is $v(a)=\frac{-2}{a^3}$ (m/s)
The velocity of the particle at time $t=1$ is $v_1=-2$ (m/s)
The velocity of the particle at time $t=2$ is $v_2=\frac{-1}{4}$ (m/s)
The velocity of the particle at time $t=3$ is $v_3=\frac{-2}{27}$ (m/s)
Work Step by Step
Equation of motion $$s=f(t)=\frac{1}{t^2}$$
* Find the equation for the velocity of the particle at time $t=a$:
According to definition, the velocity of the particle at time $t=a$ is $$v(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{\frac{a^2-(a+h)^2}{a^2(a+h)^2}}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{a^2-a^2-h^2-2ah}{a^2h(a+h)^2}$$
$$v(a)=\lim\limits_{h\to0}\frac{-h^2-2ah}{a^2h(a+h)^2}$$
$$v(a)=\lim\limits_{h\to0}\frac{-h-2a}{a^2(a+h)^2}$$
$$v(a)=\frac{-0-2a}{a^2(a+0)^2}=\frac{-2a}{a^4}=\frac{-2}{a^3}\hspace{0.1cm}(m/s)$$
*At $t=1$, the velocity of the particle is: $$v_1=\frac{-2}{1^3}=-2\hspace{0.1cm}(m/s)$$
*At $t=2$, the velocity of the particle is: $$v_2=\frac{-2}{2^3}=\frac{-1}{4}\hspace{0.1cm}(m/s)$$
*At $t=3$, the velocity of the particle is: $$v_3=\frac{-2}{3^3}=\frac{-2}{27}\hspace{0.1cm}(m/s)$$