Chapter 11.4, Problem 11.92P

The motion of a particle is defined by the equations $x=10t-5\sin t$ and $y=10-5\cos t$ , where x and y are expressed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval $0\le t\le 2\pi$ , and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times, positions, and directions of the velocities.

To determine

(a)

The magnitudes of the smallest and largest velocities reached by the particle.

Answer to Problem 11.92P

$v_{\min }=5ft/s$

$v_{\max }=15ft/s$

Explanation of Solution

Given information:

The motion of a particle is defined by the equation,

$x=10t-5\sin tandy=10-5\cos t$

Where x and y is expressed in feet and t is in seconds.

Sketch the path of the particle as plotted y versus x in the graph.

By using the equation of x and y the value is obtained and shown in the table below:

The graph between x and y.

The velocity versus time curve.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, $v_x=\frac{dx}{dt}$

$\begin{array}{l}{v}_x=\frac{d}{dt}\left(10t-5\sin t\right)\\ v_x=10-5\cos t\end{array}$

And, $v_y=\frac{dy}{dt}$

$\begin{array}{l}{v}_y=\frac{d}{dt}\left(10-5\cos t\right)\\ v_y=5\sin t\end{array}$

Now, velocity:

$v^2=v_x^2+v_y^2$

$\begin{array}{l}{v}^2={\left(10-5\cos t\right)}^2+{\left(5\sin t\right)}^2\\ v^2=100+25{\cos }^{2}t-100\cos t+25{\sin }^{2}t\\ v^2=100+25\left({\sin }^{2}t+{\cos }^{2}t\right)-100\cos t\\ v^2=125-100\cos t\left[\therefore {\sin }^{2}t+{\cos }^{2}t=1\right]\end{array}$

Again, differentiating the above value of v² with respect to t:

$\begin{array}{l}\frac{d{\left(v\right)}^2}{dt}=\frac{d\left(125-100\cos t\right)}{dt}\\ \frac{d{\left(v\right)}^2}{dt}=100\sin t\end{array}$

When $t=2N\pi ,$

$\cos t=1$

Then,

$\begin{array}{l}{v}^2=125-100\cos t\\ v^2=125-100\\ v^2=25or,\\ v=5ft/s\end{array}$ and $v^2$ is minimum.

When $t=\left(2N+1\right)\pi$

$\cos t=-1$

Then,

$\begin{array}{l}{v}^2=125-100\cos t\\ v^2=125-100\left(-1\right)\\ v^2=225or,\\ v=15ft/s\end{array}$ and $v^2$ is maximum.

To determine

(b)

The corresponding times, positions and directions of the velocities.

Answer to Problem 11.92P

$x=20\pi Nft$

$y=5ft$

$v_x=5ft,$

$v_y=0$

$\theta =0$ And,

$x=20\pi \left(N+1\right)ft$

$y=15ft$

$v_x=15ft$

$v_y=0$

$\theta =0$

Explanation of Solution

Given information:

The motion of a particle is defined by the equation,

$x=10t-5\sin tandy=10-5\cos t$

Where x and y is expressed in feet and t is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, $v_x=\frac{dx}{dt}$

$\begin{array}{l}{v}_x=\frac{d}{dt}\left(10t-5\sin t\right)\\ v_x=10-5\cos t\end{array}$

And, $v_y=\frac{dy}{dt}$

$\begin{array}{l}{v}_y=\frac{d}{dt}\left(10-5\cos t\right)\\ v_y=5\sin t\end{array}$

Now, velocity

$v^2=v_x^2+v_y^2$

$\begin{array}{l}{v}^2={\left(10-5\cos t\right)}^2+{\left(5\sin t\right)}^2\\ v^2=100+25{\cos }^{2}t-100\cos t+25{\sin }^{2}t\\ v^2=100+25\left({\sin }^{2}t+{\cos }^{2}t\right)-100\cos t\\ v^2=125-100\cos t\left[\therefore {\sin }^{2}t+{\cos }^{2}t=1\right]\end{array}$

Again differentiating the above value of v² with respect to t:

$\begin{array}{l}\frac{d{\left(v\right)}^2}{dt}=\frac{d\left(125-100\cos t\right)}{dt}\\ \frac{d{\left(v\right)}^2}{dt}=100\sin t\end{array}$

When $v=v_{\min }$

When N=0, 1, 2............

Then the position of particle at x and y;

$\begin{array}{l}x=10\left(2\pi N\right)-5\sin \left(2\pi N\right)\\ x=20\pi Nft\end{array}$

$\begin{array}{l}y=10-5\cos \left(2\pi N\right)\\ y=5ft\end{array}$

And velocities:

$\begin{array}{l}{v}_x=10-5\cos \left(2\pi N\right)\\ v_x=5ft\end{array}$

$\begin{array}{l}{v}_y=5\sin \left(2\pi N\right)\\ v_y=0\end{array}$

$\begin{array}{l}\tan \theta =\frac{v_y}{v_x}\\ \tan \theta =\frac{0}{5}=0\end{array}$

When $v=v_{\max }$

When N=0, 1, 2............

Then the position of particle at x and y;

$\begin{array}{l}x=10\left(2\pi \left(N-1\right)\right)-5\sin \left(2\pi \left(N+1\right)\right)\\ x=20\pi \left(N+1\right)ft\end{array}$

$\begin{array}{l}y=10-5\cos \left(2\pi \left(N+1\right)\right)\\ y=15ft\end{array}$

And velocities:

$\begin{array}{l}{v}_x=10-5\cos \left(2\pi \left(N+1\right)\right)\\ v_x=15ft\end{array}$

$\begin{array}{l}{v}_y=5\sin \left(2\pi \left(N+1\right)\right)\\ v_y=0\end{array}$

$\begin{array}{l}\tan \theta =\frac{v_y}{v_x}\\ \tan \theta =\frac{0}{15}=0\end{array}$