Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 13

Answer

The velocity of the ball at $t=2$ is $$v=-24ft/s$$

Work Step by Step

The function of height after $t$ seconds: $$y=f(t)=40t-16t^2$$ The velocity of a ball thrown into the air: $v_i=40ft/s$ *Find the function of the change of velocity over time: According to definition, the function of the instantaneous $v(a)$ at time $t=a$ is $$v(a)=\lim\limits_{h\to0}\frac{f(t+h)-f(t)}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{40(t+h)-16(t+h)^2-(40t-16t^2)}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{(40t+40h)+(-16t^2-16h^2-32th)-40t+16t^2}{h}$$$$v(a)=\lim\limits_{h\to0}\frac{40h-16h^2-32th}{h}$$$$v(a)=\lim\limits_{h\to0}(40-16h-32t)$$$$v(a)=40-16\times0-32t$$$$v(a)=40-32t$$ Therefore, the velocity of the ball at time $t=2$ is $$v=40-32\times2=-24ft/s$$ This answer shows that within 2 seconds, the ball has reached the top and at the time $t=2$, the ball is falling down with a speed of $24ft/s$. (Since velocity is a vector, it can have negative values. The negative sign shows that the object has moved in the other direction from the initial one. Speed, however, is not a vector, so it can only have positive values.)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.