## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 43

#### 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)

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