University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 29


The object falls at a speed of $19.6m/s$ at $2$ sec after it is dropped.

Work Step by Step

The function of the height above ground of the object after $t$ seconds of falling is $f(t)=100-4.9t^2$ The falling velocity of the object at $t=2$ seconds, therefore, is the rate of change of the height $f(t)$ with respect to $t=2$, or in other words, the derivative $f'(2)$, which can be calculated by $$f'(2)=\lim_{h\to0}\frac{f(h+2)-f(2)}{h}=\lim_{h\to0}\frac{100-4.9(h+2)^2-(100-4.9\times2^2)}{h}$$ $$f'(2)=\lim_{h\to0}\frac{100-4.9(h^2+4h+4)-(100-4.9\times4)}{h}$$ $$f'(2)=\lim_{h\to0}\frac{-4.9h^2-19.6h+(100-4.9\times4)-(100-4.9\times4)}{h}$$ $$f'(2)=\lim_{h\to0}\frac{-4.9h^2-19.6h}{h}=\lim_{h\to0}(-4.9h-19.6)$$ $$f'(2)=-4.9\times0-19.6=-19.6(m/s)$$ The negative sign is due to the fact the velocity is a vector entity, not a number entity as speed; the negative sign shows that the object is falling down. So we can drop the negative sign and say that the object falls at a speed of $19.6m/s$ at $2$ sec after it is dropped.
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