Chapter 10.2, Problem 42E

To determine

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

Answer to Problem 42E

The answer: The position of the particle at time $t=3$ is equal to $\left(e^3-4.5,e^3+4.5\right)$ .

Explanation of Solution

Given information:

The velocity of $v(t)=e^t-t,e^t+t;(1,1)$

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,

  $v(t)=e^t-t,e^t+t$

Then,

  $\begin{array}{c}{v}_1(t)=e^t-t\\ v_2(t)=e^t+t\end{array}$

Since,

  

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

  $t=3$ is equal to $(e^3-5.5,e^3+3.5)$ .

The position of the particle at time $t=0$ is equal to $(1,1)$ , which means that,

  $r(0)=(1,1)$

Then,

  $\begin{array}{c}r(3)=(1,1)+(e^3-5.5,e^3+3.5)\\ =(e^3-5.5+1,e^3+3.5+1)\\ =(e^3-4.5,e^3+4.5)\end{array}$

Therefore, the required position of the particle at time $t=3$ is equal to $\left(e^3-4.5,e^3+4.5\right)$ .

$\text{(b)}$

To determine

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

  $t=3$ .

Answer to Problem 42E

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

  $t=3$ is approximately equal to $27.791$ .

Explanation of Solution

Given information:

The velocity of $v(t)=e^t-t,e^t+t;(1,1)$ .

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,

  $v(t)=e^t-t,e^t+t$

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 approximately equal to $27.791$ .