Chapter 2.5, Problem 33E

a.

To determine

To calculate: The velocity, speed , jerk and acceleration at time $t$ .

Answer to Problem 33E

The velocity is $v=2\cos t-3\sin t$ , acceleration is $a=-2\sin t-3\cos t$ and $Speed=|2\cos t-3\sin t|$

Explanation of Solution

Given information:

The function is. $s=2-2\sin t$

Concept used:

The formula used is $\frac{d}{dx}\left(\sin x\right)=\cos x$ , $\frac{d}{dx}\left(\cos x\right)=-\sin x$ .

Calculation:

The function is $s=2-2\sin t$ .

Differentiate the function with respect to time $t$ , to find the velocity.

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(2-2\sin t\right)\\ =\frac{d}{dt}\left(2\right)-\frac{d}{dt}\left(2\sin t\right)\\ =0-2\cos t\\ v=-2\cos t\end{array}$

The speed is always a positive quantity.

  $\begin{array}{c}Speed=|-2\cos t|\\ =2\cos t\end{array}$

Differentiate the function $v=-2\cos t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(-2\cos t\right)=-2\frac{d}{dt}\left(\cos t\right)\\ a=2\sin t\end{array}$

Differentiate the function $a=2\sin t$ with respect to time $t$ , to find the jerk.

  $\begin{array}{c}\frac{d}{dt}\left(2\sin t\right)=2\frac{d}{dt}\left(\sin t\right)\\ jerk=2\cos t\end{array}$

b.

To determine

To calculate: The velocity, speed and acceleration at time $t=\frac{\pi }{4}$ .

Answer to Problem 33E

The velocity is $v=-\frac{4}{\sqrt{2}}$ and acceleration is $a=\frac{1}{\sqrt{2}}$ and $Speed=\frac{4}{\sqrt{2}}$ .

Explanation of Solution

Given information:

The function is. $s=2-2\sin t$

Concept used:

The formula used is $\frac{d}{dx}\left(\sin x\right)=\cos x$ , $\frac{d}{dx}\left(\cos x\right)=-\sin x$ .

Calculation:

The function is $s=2-2\sin t$ .

Differentiate the function with respect to time $t$ , to find the velocity.

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(2-2\sin t\right)\\ =\frac{d}{dt}\left(2\right)-\frac{d}{dt}\left(2\sin t\right)\\ =0-2\cos t\\ v=-2\cos t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the velocity $v=-2\cos t$ .

  $\begin{array}{l}v=-2\cos \frac{\pi }{4}\\ v=-\frac{2}{\sqrt{2}}\\ v=-\sqrt{2}m/\sec \end{array}$

The speed is always a positive quantity.

  $Speed=|-2\cos t|$

Substitute $\frac{\pi }{4}$ for $t$ in the $Speed=|-2\cos t|$ .

  $\begin{array}{c}Speed=\left|\left.-2\cos \frac{\pi }{4}\right|\right.\\ =\left|\left.-\frac{2}{\sqrt{2}}\right|\right.\\ =\sqrt{2}m/s\end{array}$

Differentiate the function $v=-2\cos t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(-2\cos t\right)=-2\frac{d}{dt}\left(\cos t\right)\\ a=2\sin t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the acceleration $a=2\sin t$ .

  $\begin{array}{c}a=2\sin \frac{\pi }{4}\\ =\sqrt{2}m/{\sec }^2\end{array}$

Differentiate the function $a=2\sin t$ with respect to time $t$ , to find the jerk.

  $\begin{array}{c}\frac{d}{dt}\left(2\sin t\right)=2\frac{d}{dt}\left(\sin t\right)\\ jerk=2\cos t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the acceleration $jerk=2\cos t$ .

  $\begin{array}{c}jerk=2\cos \frac{\pi }{4}\\ =2\times \frac{1}{\sqrt{2}}\\ =\sqrt{2}m/{\sec }^3\end{array}$

c.

To determine

To calculate: The motion of the particle.

Explanation of Solution

Given information:

The function is. $s=2-2\sin t$

Concept used:

The standard form of the motion is $y=a+b\sin t$ .

Calculation:

The particle will start its motion at $t=0$

Substitute $t=0$ in $s=2-2\sin t$ .

  $\begin{array}{c}s=2-2\sin t\\ =2-2\sin \left(0\right)\\ =2-0\\ =2\end{array}$

The mean position is $2$ and the amplitude is $-2$ so the particle will go up $2+\left(-2\right)=0$ and will go down $2-\left(-2\right)=4$

The period of motion is $2\pi$ .

The greatest velocity of the body can be achieved when it is either $-1$ or $1$ for that the value of $t$ should be of the form $k\pi$ and the velocity is $0$ when $t$ is of the form $k\frac{\pi }{2},k\text{ is odd}$

The acceleration is greatest when $t$ is either $-1$ or $1$ for that the value of $t$ should of the form $k\frac{\pi }{2},k\text{ is odd}$ and $0$ when $t$ is of the form $k\pi$ .

The greatest jerk of the body can be achieved when it is either $-1$ or $1$ for that the value of $t$ should be of the form $k\pi$ and the velocity is $0$ when $t$ is of the form $k\frac{\pi }{2},k\text{ is odd}$