#### Answer

The velocity and speed of the particle are both $32m/s$.

#### Work Step by Step

Equation of motion $$f(t)=80t-6t^2$$
According to definition, the velocity of the particle at time $t=a$ is $$v(a)=f'(t)=\lim\limits_{h\to0}\frac{f(t+h)-f(t)}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{[80(t+h)-6(t+h)^2]-[80t-6t^2]}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{(80t+80h-6t^2-6h^2-12th)-80t+6t^2}{h}$$
$$v(a)=\lim\limits_{h\to0}\frac{80h-6h^2-12th}{h}$$
$$v(a)=\lim\limits_{h\to0}(80-6h-12t)$$
$$v(a)=80-6\times0-12t$$
$$v(a)=80-12t\hspace{0.1cm}(m/s)$$
When $t=4$, the velocity of the particle is $$v=80-12\times4=32\hspace{0.1cm}(m/s)$$
The speed of the particle therefore is also $32m/s$
(The speed of the particle equals $|v|$ of the particle. Since speed is not a vector, it cannot be negative like velocity)