Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 13E from Chapter 2.7 from Stewart's Calculus, 8th Edition.
Problem 13E
Chapter:
Problem:
If a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y − 40t...
Step-by-Step Solution
Given information
We are given that a ball is thrown with a velocity 40 ft/s. The height of the ball after $t$ seconds is given by\[y = 40t - 16{t^2}\]We have to find the velocity of the ball at t=2 seconds.
Step 1:
The velocity of ball at any instant $t=a $ seconds is given by:
\[\begin{array}{l}v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {40 \times (2 + h) - 16 \times {{(2 + h)}^2}} \right) - \left( {40 \times 2 - 16 \times {2^2}} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {80 + 40h - 16\left( {4 + {h^2} + 4h} \right)} \right) - \left( {80 - 64} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{80 + 40h - 64 - 16{h^2} - 64h - 16}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 16{h^2} - 24h}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( { - 16h - 24} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \left( { - 16h - 24} \right)\\ = - 24\end{array}\]Hence velocity is constant and is equal to \[v\left( 2 \right) = - 24\,\,{\rm{ft/s}}\]
We are given that a ball is thrown with a velocity 40 ft/s. The height of the ball after $t$ seconds is given by\[y = 40t - 16{t^2}\]We have to find the velocity of the ball at t=2 seconds.
Step 1:
The velocity of ball at any instant $t=a $ seconds is given by:
\[\begin{array}{l}v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {40 \times (2 + h) - 16 \times {{(2 + h)}^2}} \right) - \left( {40 \times 2 - 16 \times {2^2}} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {80 + 40h - 16\left( {4 + {h^2} + 4h} \right)} \right) - \left( {80 - 64} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{80 + 40h - 64 - 16{h^2} - 64h - 16}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 16{h^2} - 24h}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left( { - 16h - 24} \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \left( { - 16h - 24} \right)\\ = - 24\end{array}\]Hence velocity is constant and is equal to \[v\left( 2 \right) = - 24\,\,{\rm{ft/s}}\]