Chapter 11.4, Problem 11.93P

To determine

(a)

The position, velocity and acceleration at time is equal to $0\text{s}$.

Answer to Problem 11.93P

The position, velocity and acceleration at time $t=0\text{s}$ are $20\text{mm}$, $43.3\text{mm/s,}$ and $743{\text{mm/s}}^{\text{2}}$ respectively.

Explanation of Solution

Given:

Position of the particle is $r=x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j$. Position $x_1=30\text{mm}$ and $y_1=20\text{mm}$.

Concept used:

Relation between position and velocity is given as:

$v=\frac{dr}{dt}$ ...... (1)

Here, time is $t$, position of the particle is $x$ and velocity of the particle is $v$.

Relation between acceleration and velocity is given as:

$a=\frac{dv}{dt}$ ...... (2)

Here, acceleration of the particle is $a$, time is $t$ and velocity of the particle is $v$.

Calculation:

Substitute $0\text{s}$ for $t$ in the position equation.

$\begin{array}{l}r=x_1\left(1-\left(\frac{1}{0+1}\right)\right)i+y_1{e}^{\frac{-\pi \times 0}{2}}\left(\cos 2\pi 0\right)j\\ r=0i+20j\end{array}$

The magnitude is calculated as follows:

$\begin{array}{l}r=\sqrt{0^2+{20}^2}\\ r=20\text{mm}\end{array}$

Substitute $x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j$ for $r$ in equation (1).

$\begin{array}{c}v=\frac{d\left(x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j\right)}{dt}\\ v=x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)j\end{array}$

Substitute $0\text{s}$ for $t$ in the above velocity equation.

$\begin{array}{l}v=30\left(0-\frac{0-1}{{\left(0+1\right)}^2}\right)i+20e^{\frac{-\pi \times 0}{2}}\left(-2\pi \sin 2\pi 0-\frac{\pi }{2}\cos 2\pi 0\right)j\\ v=30i-31.4159j\\ \left|v\right|=\sqrt{{30}^2+{\left(-31.4159\right)}^2}\\ \left|v\right|=43.3\text{mm/s}\end{array}$

Substitute $x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)j$ for $v$ in equation (2).

$\begin{array}{l}a=\frac{d\left(x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)\right)j}{dt}\\ a=x_1\left(\frac{0-2(t+1)}{{\left(t+1\right)}^4}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\frac{\pi }{2}\left(-2\pi \sin 2\pi t+\frac{\pi }{2}\cos 2\pi t\right)+\left({\pi }^2\sin 2\pi t-{\left(2\pi \right)}^2\cos 2\pi t\right)\right)\end{array}$

Substitute $0\text{s}$ for $t$ in the above acceleration equation.

$\begin{array}{l}a=30\left(\frac{0-2(0+1)}{{\left(0+1\right)}^4}\right)i+20e^{\frac{-\pi \times 0}{2}}\left(\frac{\pi }{2}\left(-2\pi \sin 2\pi 0+\frac{\pi }{2}\cos 2\pi 0\right)+\left({\pi }^2\sin 2\pi 0-{\left(2\pi \right)}^2\cos 2\pi 0\right)\right)j\\ a=-60i+20\left({\left(\frac{\pi }{2}\right)}^2-{\left(2\pi \right)}^2\right)j\\ a=-60i-740.22j\\ \left|a\right|=\sqrt{{\left(-60\right)}^2+{\left(-740.22\right)}^2}\\ \left|a\right|\approx 743{\text{mm/s}}^{\text{2}}\end{array}$

Thus, the position, velocity and acceleration at time $t=0\text{s}$ are $20\text{mm}$, $43.3\text{mm/s}$ and $743{\text{mm/s}}^{\text{2}}$ respectively.

Conclusion:

The position, velocity and acceleration at time $t=0\text{s}$ are $20\text{mm}$, $43.3\text{mm/s}$ and $743{\text{mm/s}}^{\text{2}}$ respectively.

To determine

(b)

The position, velocity and acceleration.

Answer to Problem 11.93P

The position, velocity and acceleration at time $t=0\text{s}$ are $18.1\text{mm}$, $5.66\text{mm/s}$ and $70.3{\text{mm/s}}^{\text{2}}$ respectively.

Explanation of Solution

Given:

Position of the particle is $r=x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j$. Position $x_1=30\text{mm}$ and $y_1=20\text{mm,}$ t = $1.5\text{s}$.

Calculation:

Substitute $1.5\text{s}$ for $t$ in the position equation.

$\begin{array}{l}r=x_1\left(1-\left(\frac{1}{1.5+1}\right)\right)i+y_1{e}^{\frac{-\pi \times 1.5}{2}}\left(\cos 2\pi \times 1.5\right)j\\ r=18i+1.9j\end{array}$

The magnitude is calculated as follows:

$\begin{array}{l}r=\sqrt{{18}^2+{1.9}^2}\\ r=18.1\text{mm}\end{array}$

Substitute $x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j$ for $r$ in equation (1).

$\begin{array}{c}v=\frac{d\left(x_1\left(1-\left(\frac{1}{t+1}\right)\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\cos 2\pi t\right)j\right)}{dt}\\ v=x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)j\end{array}$

Substitute $1.5\text{s}$ for $t$ in the above velocity equation.

$\begin{array}{l}v=30\left(0-\frac{0-1}{{\left(1.5+1\right)}^2}\right)i+20e^{\frac{-\pi \times 1.5}{2}}\left(-2\pi \sin 2\pi \times 1.5-\frac{\pi }{2}\cos 2\pi \times 1.5\right)j\\ v=4.8i-3j\\ \left|v\right|=\sqrt{{4.8}^2+{\left(3\right)}^2}\\ \left|v\right|=5.66\text{mm/s}\end{array}$

Substitute $x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)j$ for $v$ in equation (2).

$\begin{array}{l}a=\frac{d\left(x_1\left(0-\frac{0-1}{{\left(t+1\right)}^2}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(-2\pi \sin 2\pi t-\frac{\pi }{2}\cos 2\pi t\right)\right)j}{dt}\\ a=x_1\left(\frac{0-2(t+1)}{{\left(t+1\right)}^4}\right)i+y_1{e}^{\frac{-\pi t}{2}}\left(\frac{\pi }{2}\left(-2\pi \sin 2\pi t+\frac{\pi }{2}\cos 2\pi t\right)+\left({\pi }^2\sin 2\pi t-{\left(2\pi \right)}^2\cos 2\pi t\right)\right)\end{array}$

Substitute $1.5\text{s}$ for $t$ in the above acceleration equation.

$\begin{array}{l}a=30\left(\frac{0-2(0+1)}{{\left(0+1\right)}^4}\right)i+20e^{\frac{-\pi \times 0}{2}}\left(\frac{\pi }{2}\left(-2\pi \sin 2\pi 0+\frac{\pi }{2}\cos 2\pi 0\right)+\left({\pi }^2\sin 2\pi 0-{\left(2\pi \right)}^2\cos 2\pi 0\right)\right)j\\ a=-3.84i+70.2j\\ \left|a\right|=\sqrt{{\left(-3.84\right)}^2+{\left(70.2\right)}^2}\\ \left|a\right|\approx 70.3{\text{mm/s}}^{\text{2}}\end{array}$

Thus, the position, velocity and acceleration at time $t=0\text{s}$ are $18.1\text{mm}$, $5.66\text{mm/s}$ and $70.3{\text{mm/s}}^{\text{2}}$ respectively.

Conclusion:

The position, velocity and acceleration at $t=1.5\text{s}$ are $18.1\text{mm}$, $5.66\text{mm/s}$ and $70.3{\text{mm/s}}^{\text{2}}$ respectively.