A particle is moving in RR^3 so that its acceleration function is a(t)=⟨2t,1,0⟩. Find the position function, r(t) of the particle if it starts at the point (−5,0,2) with initial velocity v(t)=⟨3,1,−1⟩.

Patricia Bean

Patricia Bean

Answered question

2022-07-15

A particle is moving in R 3 so that its acceleration function is a ( t ) = 2 t , 1 , 0 . Find the position function, r(t) of the particle if it starts at the point ( 5 , 0 , 2 ) with initial velocity v ( t ) = 3 , 1 , 1
We have v ( t ) = 3 , 1 , 1 so r ( t ) = v ( t ) d t= 3 , 1 , 1 d t which is 3 t , t , t
From here I don't know what to do.

Answer & Explanation

lelapem

lelapem

Beginner2022-07-16Added 12 answers

Let
a ( t ) = 2 t , 1 , 0
According to the definition
d v d t = a
Integrating with respect to time
v ( t ) v ( 0 ) = 0 t a ( λ ) d λ
So we have
v ( t ) = 3 , 1 , 1 + t 2 , t , 0 = t 2 + 3 , t + 1 , 1
Similarly
r ( t ) r ( 0 ) = 0 t v ( λ ) d λ
And then
r ( t ) = 5 , 0 , 2 + t 3 / 3 + 3 t , t 2 / 2 + t , t = t 3 / 3 + 3 t 5 , t 2 / 2 + t , t + 2

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