## Calculus 8th Edition

Published by Cengage

# Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 13

#### Answer

$v(2) =-24$ft/s

#### Work Step by Step

Given the function $s(t) = 40t - 16t^2$ You can get the velocity by evaluating the derivative at $t = 2$ $v(2) = \lim\limits_{t \to 2}\frac{s(t) - s(2)}{t-2}$ Substitute $s(t)$ and $s(2) (= 80 - 64 = 16)$ $v(2) = \lim\limits_{t \to 2}\frac{(40t - 16t^2) - 16}{t-2}$ Factor the top $v(2) = \lim\limits_{t \to 2}\frac{-8(2t^2-5t+2)}{t-2}$ $v(2) = \lim\limits_{t \to 2}\frac{-8(t-2)(2t-1)}{t-2}$ Simplify the expression ($(t-2)$ cancels) $v(2) = \lim\limits_{t \to 2}-8(2t-1)$ $v(2) =-8 \lim\limits_{t \to 2}(2t-1)$ Find the limit $v(2) =-8 \lim\limits_{t \to 2}(2(2)-1)$ $v(2) =-8 \lim\limits_{t \to 2}(3)$ $v(2) =-8 (3)$ $v(2) =-24$ The instantaneous velocity at 2 seconds is $-24$ft/s

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