Chapter 10.2, Problem 41E

$\text{(a)}$

To determine

To find: The position of the particle at time $t=3$ .

Answer to Problem 41E

The answer: The position of the particle at time $t=3$ is equal to $(2\ln (2)+3,-1.7)$ .

Explanation of Solution

Given information:

The velocity of $v(t)={(t+1)}^{-1},{(t+2)}^{-2};(3,-2)$

Calculation:

Know that the position vector of a particle at time $b$ is equal to the position vector of a particle at time $a$ plus the displacement of a particle from $t=a$

To

  $t=b$ .

Know that,

  $v(t)=v_1(t),v_2(t)$

Then,

  

In this exercise,

  $v(t)={(t+1)}^{-1},{(t+2)}^{-2}$

Then,

  $\begin{array}{}$ v_1(t)={(t+1)}^{-1}$$ v_2(t)={(t+2)}^{-2}$\end{array}$

Since,

  

To calculate the integral:

  

To calculate the integral:

  

To use the previous results, the displacement of the particle from $t=0$ to

  $t=3$ is equal to $(2\ln (2),0.3)$

The position of the particle at time $t=0$ is equal to $(3,-2)$ , which means that,

  $t(0)=(3,-2)$

Then,

  $\begin{array}{c}r(3)=(3,-2)+(2\ln (2),0.3)\\ =(2\ln (2)+3,-2+0.3)\\ =(2\ln (2)+3,-1.7)\end{array}$

Therefore, the required position of the particle at time $t=3$ is equal to $(2\ln (2)+3,-1.7)$ .

$\text{(b)}$

To determine

To find: The distance the particle travels from $t=0$ to

  $t=3$ .

Answer to Problem 41E

The answer: The distance the particle travels from $t=0$ to

  $t=3$ is equal to $1.419$ .

Explanation of Solution

Given information:

The velocity of $v(t)={(t+1)}^{-1},{(t+2)}^{-2};(3,-2)$

Calculation:

Know that the position vector of a particle at time $b$ is equal to the position vector of a particle at time $a$ plus the displacement of a particle from $t=a$

To

  $t=b$ .

Know that,

  $v(t)=v_1(t),v_2(t)$

Is

  

In this exercise,

  $v(t)={(t+1)}^{-1},{(t+2)}^{-2}$

Since, The distance the particle travels from $t=0$ to

  $t=3$ .

Is equal to:

  

Therefore, the required distance the particle travels from $t=0$ to

  $t=3$ is equal to $1.419$ .