   Chapter 3.9, Problem 55E

Chapter
Section
Textbook Problem

# 53-58 A particle is moving with the given data. Find the position of the particle. a ( t ) = 2 t + 1 ,    s ( 0 ) = 3 ,    v ( 0 ) = − 2

To determine

To find:

The position of the particle s(t)

Explanation

1) Concept:

i. If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is Fx+c, where c is an arbitrary constant.

ii.

at=ddxvt

iii.

vt=ddxst

2) Given:

at=2t+1,  s0=3,  v0=-2.

3) Calculations:

Here, at=2t+1

vt is the antiderivative of at

The general antiderivative of at by using rules of antiderivative is

vt=2t22+t+C

vt=t2+t+C

It is given that v0=-2

So, substitute t=0 in v(t)

v0=(0)2+(0)+C

-2=0+C

C=-2

Substitute C=-2 in vt

vt=t2+t-2

The particle is moving with a velocity of vt=t2+t-2

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