   Chapter 13, Problem 19RE

Chapter
Section
Textbook Problem

# A particle starts at the origin with initial velocity i − j + 3 k. Its acceleration is a(t) = 6t i + 12t2j − 6t k. Find its position function.

To determine

To find: The position function of a particle.

Explanation

Given data:

The particle starts at the origin with the initial velocity v(0)=ij+3k and its acceleration is a(t)=6ti+12t2j6tk .

Formula used:

Write the expression to find the acceleration.

a(t)=v(t) (1)

Write the expression to find the velocity.

v(t)=r(t) (2)

Modify equation (1), to obtain the velocity.

v(t)=a(t)

Substitute (6ti+12t2j6tk) for a(t) ,

v(t)=(6ti+12t2j6tk)dt=[(6t)dt]i+[(12t2)dt]j+[(6t)dt]k=[6(t22)]i+[12(t33)]j+[(6)(t22)]k+C=(3t2)i+(4t3)j+(3t2)k+C

v(t)=3t2i+4t3j3t2k+C (3)

Here,

C is the constant.

As the initial velocity v(0) is (ij+3k) , substitute 0 for t in equation (3) and equate the resultant with the vector (ij+3k) to obtain the value of constant C .

Substitute 0 for t in equation (3),

v(0)=3(0)2i+4(0)3j3(0)2k+C=(0)i+(0)j(0)k+C=C

v(0)=C .

Substitute (ij+3k) for v(0) ,

C=ij+3k

Substitute (ij+3k) for C in equation (3),

v(t)=3t2i+4t3j3t2k+(ij+3k)=(3t2+1)i+(4t31)j+(33t2)k

Modify equation (2), to obtain the position vector

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