Chapter 2, Problem 116P

(a)

To determine

The velocity of the particle when its position is $x=3.0\text{m}$ .

Answer to Problem 116P

The velocity of particle when its position is $x=3.0\text{m}$ is $4\text{m}/\text{s}$ .

Explanation of Solution

Given:

The acceleration of the particle is given by $a_x\left(x\right)=\left(2.0{\text{s}}^{-2}\right)x$ .

The velocity is $0\text{m}/\text{s}$ at $x=1.0\text{m}$ .

Formula used:

Write the expression for the acceleration of the particle.

  $a=\frac{dv}{dt}$

Here, $a$ is the acceleration of the particle and $\frac{dv}{dx}$ is the rate of change velocity.

Multiply by $\frac{dx}{dx}$ in the above expression.

  $\begin{array}{c}a=\frac{dv}{dt}\cdot \frac{dx}{dx}\\ =\frac{dv}{dx}\cdot \frac{dx}{dt}\end{array}$

Substitute $v$ for $\frac{dx}{dt}$ and rearrange the above expression.

  $\begin{array}{c}a=v\frac{dv}{dx}\\ vdv=adx\end{array}$

Substitute, $\left(2.0{\text{s}}^{-2}\right)x$ for $a$ in above expression and integrate for $v=0\text{m}/\text{s}$ to $v=v\text{m}/\text{s}$ and $x=x_0$ to $x=x$ .

  $\begin{array}{c}{\displaystyle \underset{v=0}{\overset{v}{\int }}vdv}={\displaystyle \underset{x_0}{\overset{x}{\int }}\left(2.0{\text{s}}^{-2}\right)x\cdot dx}\\ {\left[\frac{v^2}{2}\right]}_0^v=\left(2.0{\text{s}}^{-2}\right){\left[\frac{x^2}{2}\right]}_{x_0}^x\end{array}$

Simplify above expressinon.

  $v^2=\left(2.0{\text{s}}^{-2}\right)\left(x^2-x_0^2\right)$........(1)

Calculation:

Substitute $3.0\text{m}$ for $x$ and $1.0\text{m}$ in equation (1).

  $\begin{array}{l}{v}^2=\left(2.0{\text{s}}^{-2}\right)\left[{\left(3.0\right)}^2-{\left(1.0\right)}^2\right]\\ v^2=16\\ v=4\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of particle when its position is $x=3.0\text{m}$ is $4\text{m}/\text{s}$ .

(b)

To determine

The time taken by the particle to travel from $x=1.0\text{m}$ to $x=3.0\text{m}$ .

Answer to Problem 116P

Time taken by the particle to travel from $x=1.0\text{m}$ to $x=3.0\text{m}$ is $1.22\text{s}$ .

Explanation of Solution

Given:

The acceleration of the particle is given by $a_x\left(x\right)=\left(2.0{\text{s}}^{-2}\right)x$ .

The velocity is $0\text{m}/\text{s}$ at $x=1.0\text{m}$ .

Formula used:

Write the expression for the velocity of the particle.

  $v^2=\left(2.0{\text{s}}^{-2}\right)\left(x^2-x_0^2\right)$

Further, simplify for $v$ .

  $v=\sqrt{\left(2.0{\text{s}}^{-2}\right)\left(x^2-x_0^2\right)}$  ........(1)

Write the expression for the instantaneous velocity of the particle.

  $v=\frac{dx}{dt}$

Rearrange the above expression for $dt$ .

  $dt=\frac{dx}{v}$........(2)

Calculation:

Substitute $\sqrt{\left(2.0{\text{s}}^{-2}\right)\left(x^2-x_0^2\right)}$ for $v$ in equation (2) and integrate for $t=0\text{s}$ to $t=t\text{s}$ and $x=x_0$ to $x=x$ .

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{t}{\int }}dt}={\displaystyle \underset{x_0}{\overset{x}{\int }}\frac{dx}{\sqrt{\left(2.0{\text{s}}^{-2}\right)\left(x^2-x_0^2\right)}}}\\ {\displaystyle \underset{0}{\overset{t}{\int }}dt}=\frac{1}{\sqrt{2.0{\text{s}}^{\text{-2}}}}{\displaystyle \underset{x_0}{\overset{x}{\int }}\frac{dx}{\sqrt{\left(x^2-x_0^2\right)}}}\\ t=\frac{1}{\sqrt{2.0{\text{s}}^{\text{-2}}}}\ln \left(\frac{x+\sqrt{x^2-x_0^2}}{x_0}\right)\end{array}$

Substitute $3,0\text{m}$ for $x$ and $1.0\text{m}$ for $x_o$ in above expression.

  $\begin{array}{c}t=\frac{1}{\sqrt{2.0{\text{s}}^{\text{-2}}}}\ln \left(\frac{3+\sqrt{{\left(3\right)}^2-{\left(1\right)}^2}}{\left(1\right)}\right)\\ =\frac{1}{\sqrt{2}}\ln \left(3+2.82\right)\\ =1.22\text{s}\end{array}$

Conclusion:

Thus, time taken by the particle to travel from $x=1.0\text{m}$ to $x=3.0\text{m}$ is $1.22\text{s}$ .