Chapter 2.5, Problem 16E

a.

To determine

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

Answer to Problem 16E

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. $s=\cos t-3\sin t$

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 $s=\cos t-3\sin t$ .

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

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

The speed is always a positive quantity.

  $\begin{array}{c}Speed=|-\sin t-3\cos t|\\ =\sin t+3\cos t\end{array}$

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

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

b.

To determine

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

Answer to Problem 16E

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

Explanation of Solution

Given information:

The function is. $s=\cos t-3\sin t$

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 $s=\cos t-3\sin t$ .

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

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

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

  $\begin{array}{l}v=-\sin \frac{\pi }{4}-3\cos \frac{\pi }{4}\\ v=-\frac{1}{\sqrt{2}}-\frac{3}{\sqrt{2}}\\ v=-\frac{4}{\sqrt{2}}\end{array}$

The speed is always a positive quantity.

  $Speed=|-\sin t-3\cos t|$

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

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

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

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

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

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

c.

To determine

To calculate: The motion of the particle.

Explanation of Solution

Given information:

The function is. $s=\cos t-3\sin t$ .

Concept used:

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

Calculation:

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

Substitute $t=0$ in $s=\cos t-3\sin t$ .

  $\begin{array}{c}s=\cos t-3\sin t\\ =\cos \left(0\right)-3\sin \left(0\right)\\ =1-3\times 0\\ s=1\end{array}$

The position of the body $s=\cos t-3\sin t$ can also be written s shown below.

  $\begin{array}{c}s=\sqrt{5}\left(\frac{1}{\sqrt{5}}\cos t-\frac{3}{\sqrt{5}}\sin t\right)\\ =\sqrt{5}\left(\cos t\sin \theta -\sin t\cos \theta \right)\\ s=\sqrt{5}\left(\sin \left(\theta -wt\right)\right)\end{array}$

On comparing with the standard form the amplitude is $\sqrt{5}$ .

So it will go $\sqrt{5}$ upwards and $\sqrt{5}$ downwards.

The period of motion is $2\pi$ .