Question

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. a(t) = 2 i + 6t j + 12t2 k, v(0) = i, r(0) = 2 j − 3 k v(t) = r(t) =

Answer 1

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

When $t=0$, $v(0) = \mathbb{i}$. Inserting these values in $v(t)$,

$C_1= \mathbb{i}$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + \mathbb{i}$

$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

$r(t) = (t^2+t) \mathbb{i}+t^3 \mathbb{j} t^4 \mathbb{k} + C_2$

When $t=0$, $r(t) =2\mathbb{j}-3\mathbb{k}$. Inserting these values in $r(t)$,

$C_2=2\mathbb{j}-3\mathbb{k}$

$r(t)= (t^2+t) \mathbb{i}+t^3 \mathbb{j}+ t^4 \mathbb{k} + 2\mathbb{j}-3\mathbb{k}$

$r(t) = (t^2+t) \mathbb{i}+(t^3+2) \mathbb{j}+ (t^4-3) \mathbb{k}$