Chapter 11.1, Problem 11.19P

Based on experimental observations, the acceleration of a particle is defined by the relation $a=-\left(0.1+\sin x/b\right)$ , where a and x are expressed in m/s² and meters, respectively. Knowing that $b=0.8$ m and that $v=1$ m/s when $x=0$ , determine (a) the velocity of the particle when $x=-1$ m, (b) the position where the velocity is maximum, (c) the maximum velocity.

To determine

(a)

The velocity of particle when x = -1 m.

Answer to Problem 11.19P

Velocity $v=0.323m/\sec$ .

Explanation of Solution

Given information:

The acceleration of the particle is given by the equation:

$a=-\left(0.1+\sin \frac{x}{b}\right)$

Where (a) in m/s² and (x) is in meters.

Also the given condition is that: v=1m/s when x=0 and b=0.8 m.

The basic kinematics relation is given by the equation, if $a=a\left(x\right)$ then,

$v\frac{dv}{dx}=a\left(x\right)$

$v\frac{dv}{dx}=-\left(0.1+\sin \frac{x}{0.8}\right)$ (Given)

Now, integrate the above equation within the limits when x=0, v=1 m/s.

$\underset{1}{\overset{v}{\int }}dv=\underset{0}{\overset{x}{\int }}-\left(0.1+\sin \frac{x}{0.8}\right)dx$

$\begin{array}{l}\frac{1}{2}{\left[v^2\right]}_1^v=-{\left[0.1x-0.8\cos \frac{x}{0.8}\right]}_0^x\\ \frac{1}{2}\left[v^2-1\right]=-\left[0.1\left(x-0\right)-0.8\left(\cos \frac{x}{0.8}-\cos \frac{0}{0.8}\right)\right]\\ \frac{1}{2}{v}^2-\frac{1}{2}=-0.1x+0.8\cos \frac{x}{0.8}-0.8\times 1\\ \frac{1}{2}{v}^2=-0.1x+0.8\cos \frac{x}{0.8}-0.8+0.5\end{array}$

$\frac{1}{2}{v}^2=-0.1x+0.8\cos \frac{x}{0.8}-0.3$

Put x=-1 in the equation, we get:

$\frac{1}{2}{v}^2=-0.1(-1)+0.8\cos \frac{\left(-1\right)}{0.8}-0.3$

$\begin{array}{l}\frac{1}{2}{v}^2=+0.1+0.8\cos \left(-1.25\right)-0.3\\ \frac{1}{2}{v}^2=0.3153\times 0.8-0.2\left(\cos \left(-\theta \right)=\cos \theta \right)\end{array}$

$\begin{array}{l}\frac{1}{2}{v}^2=0.2522-0.2\\ \frac{1}{2}{v}^2=0.0522\\ v^2=0.0522\times 2\\ v^2=0.10448\\ v=\sqrt{0.10448}\\ v=.323m/\sec \\ \end{array}$

Conclusion:

The velocity of the particle when x=-1 m is $v=.323m/\sec$.

To determine

(b)

The position where the velocity is maximum.

Answer to Problem 11.19P

The position is $x=-0.0801m$.

Explanation of Solution

Given information:

The motion of the particle is given by the equation:

$a=-\left(0.1+\sin \frac{x}{b}\right)$

Where (a) in m/s² and (x) is in meters.

Also the given condition is that: v=1m/s when x=0 and b=0.8 m.

When v=v_max, a=0

From above given equation:

$\begin{array}{l}a=-\left(0.1+\sin \frac{x}{0.8}\right)\\ a=-\left(0.1+\sin \frac{x}{0.8}\right)=0\end{array}$ Or,

$\begin{array}{l}\sin \frac{x}{0.8}=-0.1\\ \frac{x}{0.8}={\sin }^{-1}(-0.1)\\ \frac{x}{0.8}=-0.10016\\ x=-0.10016\times 0.8\\ x=-0.080133\approx 0.0801\\ x=-0.0801m\end{array}$

Conclusion:

The position when velocity is maximum is $x=-0.0801m$.

To determine

(c)

The maximum velocity.

Answer to Problem 11.19P

The maximum velocity

Explanation of Solution

Given information:

The motion of the particle is given by the equation:

$a=-\left(0.1+\sin \frac{x}{b}\right)$

Where (a) in m/s² and (x) is in meters.

Also, the given condition is that: v=1m/s when x=0 and b=0.8 m.

The basic kinematics relation is given by the equation:

$a=v\frac{dv}{dx}$

$v\frac{dv}{dx}=-\left(0.1+\sin \frac{x}{0.8}\right)$ (Given)

Now, integrate the above equation within the limits when x=0, v=1 m/s.

$\underset{1}{\overset{v}{\int }}dv=\underset{0}{\overset{x}{\int }}-\left(0.1+\sin \frac{x}{0.8}\right)dx$

$\begin{array}{l}\frac{1}{2}{\left[v^2\right]}_1^v=-{\left[0.1x-0.8\cos \frac{x}{0.8}\right]}_0^x\\ \frac{1}{2}\left[v^2-1\right]=-\left[0.1\left(x-0\right)-0.8\left(\cos \frac{x}{0.8}-\cos \frac{0}{0.8}\right)\right]\\ \frac{1}{2}{v}^2-\frac{1}{2}=-0.1x+0.8\cos \frac{x}{0.8}-0.8\times 1\\ \frac{1}{2}{v}^2=-0.1x+0.8\cos \frac{x}{0.8}-0.8+0.5\end{array}$

$\frac{1}{2}{v}^2=-0.1x+0.8\cos \frac{x}{0.8}-0.3$

Put x=-0.0801 in the equation, we get:

$\begin{array}{l}\frac{1}{2}{v}^2=-0.1\left(-0.0801\right)+0.8\cos \frac{-0.0801}{0.8}-0.3\\ \frac{1}{2}{v}^2=0.0080134+0.8\cos \left(.100125\right)-0.3\\ \frac{1}{2}{v}^2=0.0080134+0.8\times 0.99-0.3\\ \frac{1}{2}{v}^2=0.0080134+0.792-0.3\\ \frac{1}{2}{v}^2=0.5032\\ v^2=1.006\\ v=\sqrt{1.006}\\ v=1.003m/\sec \\ \end{array}$

Conclusion:

The velocity of the particle when x=-0.0801 m is $v=1.003m/\sec$.