Chapter 2.5, Problem 34E

a.

To determine

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

Answer to Problem 34E

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

Explanation of Solution

Given information:

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

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

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

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

The speed is always a positive quantity.

  $Speed=|\cos t-\sin t|$

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

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

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

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

b.

To determine

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

Answer to Problem 34E

The velocity is $v=0m/\sec$

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

  $a=\frac{1}{\sqrt{2}}$ and $Speed=0m/s$ , jerk is $0m/{\sec }^3$

Explanation of Solution

Given information:

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

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

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

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

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

  $\begin{array}{l}v=\cos \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{4}\right)\\ v=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\ v=0m/\sec \end{array}$

The speed is always a positive quantity.

  $Speed=|\cos t-\sin t|$

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

  $\begin{array}{c}Speed=\left|\cos \left(\frac{\pi }{4}\right)-\left.\sin \left(\frac{\pi }{4}\right)\right|\right.\\ =\left|\left.\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right|\right.\\ Speed=0m/s\end{array}$

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

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

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

  $\begin{array}{c}a=-\sin \frac{\pi }{4}-\cos \frac{\pi }{4}\\ =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\ =-\frac{2}{\sqrt{2}}\\ a=-\frac{1}{\sqrt{2}}m/{\sec }^2\end{array}$

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

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

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

  $\begin{array}{c}jerk=-\cos \left(\frac{\pi }{4}\right)+\sin \left(\frac{\pi }{4}\right)\\ =-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ =0m/{\sec }^3\end{array}$

c.

To determine

To calculate: The motion of the particle.

Explanation of Solution

Given information:

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

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

  $\begin{array}{c}s=\sin t+\cos t\\ =0+1\\ =1\end{array}$

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

  $\begin{array}{c}s=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin t+\frac{1}{\sqrt{2}}\cos t\right)\\ =\sqrt{2}\left(\cos \theta \sin t+\sin \theta \cos t\right)\\ s=\sqrt{2}\left(\sin \left(\theta +wt\right)\right)\end{array}$

On comparing with the standard form the amplitude is $\sqrt{2}$ .

So it will go $\sqrt{2}$ upwards and $\sqrt{2}$ downwards.

The period of motion is $2\pi$ .