Chapter 2.5, Problem 15E

a.

To determine

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

Answer to Problem 15E

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. $s=2\sin t+3\cos t$

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 $s=2\sin t+3\cos t$ .

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

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

The speed is always a positive quantity.

  $Speed=|2\cos t-3\sin t|$

Differentiate the function $v=4\sin t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(2\cos t-3\sin t\right)=2\frac{d}{dt}\left(\cos t\right)-3\frac{d}{dt}\left(\sin t\right)\\ a=-2\sin t-3\cos t\end{array}$

b.

To determine

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

Answer to Problem 15E

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. $s=2\sin t+3\cos t$

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 $s=2\sin t+3\cos t$ .

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

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

Substitute $\frac{\pi }{4}$ for $t$ in the velocity $v=2\cos t-3\sin t$ .

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

The speed is always a positive quantity.

  $Speed=|2\cos t-3\sin t|$

Substitute $\frac{\pi }{4}$ for $t$ in the $Speed=|2\cos t-3\sin t|$ .

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

Differentiate the function $v=2\cos t-3\sin t$ with respect to time $t$ , to find the acceleration.

  $\begin{array}{c}\frac{d}{dt}\left(2\cos t-3\sin t\right)=2\frac{d}{dt}\left(\cos t\right)-3\frac{d}{dt}\left(\sin t\right)\\ a=-2\sin t-3\cos t\end{array}$

Substitute $\frac{\pi }{4}$ for $t$ in the acceleration $a=-2\sin t-3\cos t$ .

  $\begin{array}{c}a=-2\sin \frac{\pi }{4}-3\cos \frac{\pi }{4}\\ =-\frac{2}{\sqrt{2}}-\frac{3}{\sqrt{2}}\\ =-\frac{5}{\sqrt{2}}\end{array}$

c.

To determine

To calculate: The motion of the particle.

Explanation of Solution

Given information:

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

Concept used:

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

Calculation:

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

Substitute $t=0$ in $s=2\sin t+3\cos t$ .

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

The position of the body $s=2\sin t+3\cos t$ can also be written s shown below.

  $\begin{array}{c}s=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin t+\frac{3}{\sqrt{13}}\cos t\right)\\ =\sqrt{13}\left(\cos \theta \sin t+\sin \theta \cos t\right)\\ s=\sqrt{13}\left(\sin \left(\theta +wt\right)\right)\end{array}$

On comparing with the standard form the amplitude is $\sqrt{13}$ .

So it will go $\sqrt{13}$ upwards and $\sqrt{13}$ downwards.

The period of motion is $2\pi$ .