Chapter 2.5, Problem 13E

a.

To determine

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

Answer to Problem 13E

The velocity is $v=3\cos t$ and acceleration is $a=-3\sin t$ and $Speed=\left|3\cos t\right|$ .

Explanation of Solution

Given information:

The function is. $s=2+3\sin t$

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 $s=2+3\sin t$ .

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

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(2+3\sin t\right)\\ =\frac{d}{dt}\left(2\right)+\frac{d}{dt}\left(3\sin t\right)\\ =0+3\cos t\\ v=3\cos t\end{array}$

The speed is always a positive quantity.

  $Speed=\left|3\cos t\right|$

Differentiate the function $v=3\cos t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(3\cos t\right)=3\frac{d}{dt}\left(\cos t\right)\\ a=-3\sin t\end{array}$

b.

To determine

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

Answer to Problem 13E

The velocity is $v=3\cos t$ and acceleration is $a=-3\sin t$ and $Speed=\left|3\cos t\right|$ .

Explanation of Solution

Given information:

The function is. $s=2+3\sin t$

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 $s=2+3\sin t$ .

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

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(2+3\sin t\right)\\ =\frac{d}{dt}\left(2\right)+\frac{d}{dt}\left(3\sin t\right)\\ =0+3\cos t\\ v=3\cos t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the velocity $v=3\cos t$ .

  $\begin{array}{l}v=3\cos \frac{\pi }{4}\\ v=\frac{3}{\sqrt{2}}\end{array}$

The speed is always a positive quantity.

  $Speed=\left|3\cos t\right|$

Substitute $\frac{\pi }{4}$ for $t$ in the $Speed=\left|3\cos t\right|$ .

  $\begin{array}{c}Speed=\left|\left.3\cos \frac{\pi }{4}\right|\right.\\ =\left|\left.3\times \frac{1}{\sqrt{2}}\right|\right.\\ =\left|\left.\frac{3}{\sqrt{2}}\right|\right.\end{array}$

Differentiate the function $v=3\cos t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(3\cos t\right)=3\frac{d}{dt}\left(\cos t\right)\\ a=-3\sin t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the acceleration $a=-3\sin t$ .

  $\begin{array}{c}a=-3\sin \frac{\pi }{4}\\ =-3\times \frac{1}{\sqrt{2}}\\ =-\frac{3}{\sqrt{2}}\end{array}$

c.

To determine

To calculate: The motion of the particle.

Explanation of Solution

Given information:

The function is. $s=2+3\sin t$

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 $s=2+3\sin t$ .

  $\begin{array}{c}s=2+3\sin \left(0\right)\\ =2+0\\ s=2\end{array}$

The mean position is $2$ and the amplitude is $3$ so the particle will go up $2+3=5$ and will go down $2-3=-1$

The period of motion is $2\pi$ .