   Chapter 9.8, Problem 4CP ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Suppose that the distance a particle travels is given by s = 4 x 3 − 12 x 2 + 6 where s is in feet and x is in seconds.When does the velocity of this particle increase?

To determine

The condition when velocity of the particle increases if distance function of the particle is s=4x312x2+6, where s is in feet and x is in second.

Explanation

Given Information:

The distance function of the particle is s=4x312x2+6, where s is in feet and x is in second.

Explanation:

Consider the provided function,

s=4x312x2+6

The velocity of a particle is the derivative of the distance function of the particle.

To find the velocity function, differentiate both sides of the function s with respect to x,

dsdx=ddx(4x312x2+6)=ddx(4x3)ddx(12x2)+ddx(6)=4ddx(x3)12ddx(x2)+ddx(6)

Simplify the derivative using the power rule and the rule of derivative of a constant,

dsdx=4(3x31)12(2x21)+0=12x224x

The acceleration of a particle is the rate of change of velocity of the particle

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