   Chapter 12.3, Problem 22E

Chapter
Section
Textbook Problem

# Finding a Position Vector by Integration In Exercises 21-26, use the given acceleration vector and initial conditions to find the velocity and position vectors. Then find the position at time t = 2 . a ( t ) = − 32 k , v ( 0 ) = 3 i − 2 j + k ,     r ( 0 ) = 5 j + 2 k

To determine

To calculate: The velocity and position vector at t=0, position vector at t=2 for given acceleration vector a(t)=32k^.

Explanation

Given:

The given acceleration vector is and initial conditions a(t)=32k^; v(0)=3i^2j^+k^, r(0)=5j^+2k^

Formula used:

The integrations formula is:

v(t)=a(t)dtr(t)=v(t)dt

Calculation:

The velocity vector is v(t)=a(t)dt=32k^.dt=32tk^+c

Where

c=c1i^+c2j^+c3k^

Applying condition

v(0)=3i^-2j^+k^ we getv(0)=c1i^+c2j^+c3k^=3i^2j^+k^ c1=3, c2=2, c3=1

So the velocity at time

t is v(t)=3i^2j^+(132t)k^

By integrating we get,

r(t)=v(t)dt = 3t

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