# Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 44

The velocity of the particle is $-1.8m/s$ and the speed of the particle is $1.8m/s$ at $t=4$.

#### Work Step by Step

Equation of motion $$f(t)=10+\frac{45}{t+1}$$ The velocity of the particle at time $t=a$ can be found as follows: $$v(a)=f'(t)=\lim\limits_{t\to a}\frac{f(t)-f(a)}{t-a}$$ $$v(a)=\lim\limits_{t\to a}\frac{(10+\frac{45}{t+1})-(10+\frac{45}{a+1})}{t-a}$$ $$v(a)=\lim\limits_{t\to a}\frac{\frac{45}{t+1}-\frac{45}{a+1}}{t-a}$$ $$v(a)=\lim\limits_{t\to a}\frac{\frac{(45a+45)-(45t+45)}{(t+1)(a+1)}}{t-a}$$ $$v(a)=\lim\limits_{t\to a}\frac{45a-45t}{(t+1)(a+1)(t-a)}$$ $$v(a)=\lim\limits_{t\to a}\frac{-45(t-a)}{(t+1)(a+1)(t-a)}$$ $$v(a)=\lim\limits_{t\to a}\frac{-45}{(t+1)(a+1)}$$ $$v(a)=\frac{-45}{(a+1)(a+1)}$$ $$v(a)=\frac{-45}{(a+1)^2}$$ When $t=4$, the velocity of the particle is $$v=\frac{-45}{(4+1)^2}=\frac{-45}{25}=-1.8\hspace{0.1cm}(m/s)$$ The speed of the particle therefore is therefore $1.8m/s$ (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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