   Chapter 2.7, Problem 13E

Chapter
Section
Textbook 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 – l6t2. Find the velocity when t = 2.

To determine

To find: The velocity of the ball when t=2.

Explanation

Given:

The height (in feet) of the ball after t seconds is y=40t16t2.

The position function is, y=40t16t2.

Formula used:

The derivative f(a) is the velocity v(t) of the particle at time t=a.

v(t)=limtaf(t)f(a)ta (1)

Calculation:

Obtain the velocity of the ball when time t=2.

Take the position function y=f(t) and substitute time a=2 in equation (1)

v(t)=limt2f(t)f(2)t2=limt2(40t16t2)(40(2)16(2)2)t2=limt2(40t16t2)(8064)t2=limt240t16t216t2

=limt2(16t240t+16)t2=limt28(2t25t+2</

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