Q119P - Fundamentals Of Physics

Question

The position of a particle as it moves along a ${}^{\mathbf{y}}$ axis is given by ${}^{\mathbf{y} = \left(2.0 \mathrm{cm}\right) \sin \left(\mathbf{\pi t} /4\right)}$, with ${}^{\mathbf{t}}$ in seconds and ${}^{\mathbf{y}}$ in centimeters. (a) What is the average velocity of the particle between ${}^{\mathbf{t}\mathbf{ }\mathbf{=}\mathbf{0}}$ and ${}^{\mathbf{t}\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$? (b) What is the instantaneous velocity of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$,${}^{\mathbf{1}\mathbf{.}\mathbf{0}}$, and ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$ ?? (c) What is the average acceleration of the particle between ${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$ and ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$?  (d) What is the instantaneous acceleration of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$,${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }}$, and ${}^{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$?

Step-by-Step Solution

Answer

1. The average velocity of the particle between ${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$ and ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ is ${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$.

2. The instantaneous velocity of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ is ${}^{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$, ${}^{\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$ and ${}^{\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$ respectively.

3. The average acceleration of the particle between ${}^{\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{ }}$ and ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ is ${}^{\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$.

4. The instantaneous acceleration of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{ }}$, ${}^{\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ and ${}^{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ is ${}^{\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{,}\mathbf{-}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{87}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$ and ${}^{\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$respectively.

Step 1: Given data

The position of the particle, ${}^{\mathbf{y}\mathbf{ }\mathbf{=}\mathbf{ }\left(2.0 cm\right)\mathbf{s}\mathbf{i}\mathbf{n}\left(\frac{\mathrm{\pi t}}{4}\right)}$ 

Step 2: Understanding the average velocity, instantaneous velocity, average acceleration and instantaneous acceleration

The average velocity of an object is the ratio of total displacement to the total time.

Instantaneous velocity is the time rate of change of displacement at the given instant

Average acceleration is the change in velocity for a particular time interval whereas instantaneous acceleration is the time rate of change of velocity at a given instant of time.

The expression for the average velocity is given as follows: 

 ${\mathit{v}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\frac{\mathbf{\Delta x}}{\mathbf{\Delta t}}$                                                                                    … (i)

Here, ${}^{\mathbf{\Delta }\mathbf{x}}$ is the net displacement and ${}^{\mathbf{\Delta }\mathbf{t}}$ is the time interval.

The expression for the instantaneous velocity is given as follows: 

$\mathit{v}\mathbf{=}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}$                                                                                     … (ii)

The expression for the average acceleration is given as follows: 

$a_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{\Delta v}}{\mathbf{\Delta t}}$                                                                              … (iii)

The expression for the instantaneous acceleration is given as follows: 

$\mathit{a}\mathbf{=}\frac{\mathbf{dv}}{\mathbf{dt}}$                                                                                 … (iv)

Step 3: (a) Determination of the average velocity between t = 0 to t = 2.0 s

The position of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{s}}$ is, 

$$\begin{gathered} \mathit{y}\mathbf{=}\left(2.0cm\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi \times 0}{4}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{ }\mathit{c}\mathit{m} \end{gathered}$$

The position of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{s}}$ is, 

$$\begin{gathered} \mathit{y}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{\pi }\mathbf{\times }\mathbf{2}}{\mathbf{4}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{cm}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{c}\mathit{m} \end{gathered}$$

Using equation (i), the average velocity is calculated as follows: 

$$\begin{gathered} {\mathbf{v}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{-}\mathbf{0}\mathbf{ }\mathbf{c}\mathbf{ }\mathbf{m}}{\mathbf{2}\mathbf{ }\mathbf{s}\mathbf{-}\mathbf{0}\mathbf{ }\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s} \end{gathered}$$

Thus, the average velocity of the particle between ${}^{\mathbf{t}\mathbf{=}\mathbf{0}}$ and ${}^{\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}}$ is ${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$.

Step 4: (b) Determination of the instantaneous velocity

Using equation (ii), the instantaneous velocity is,

$$\begin{gathered} \mathbf{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left(\left(2.0 \mathrm{cm}\right)\sin \left(\frac{\mathrm{\pi t}}{4}\right)\right) \\ \mathbf{v}\mathbf{=}\left(\frac{\pi }{2}\mathrm{cm}\right)\mathbf{cos}\left(\frac{\mathrm{\pi t}}{4}\right) \end{gathered}$$

At ${}^{\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{s}}$ the instantaneous velocity is, 

 $$\begin{gathered} \mathbf{v}\mathbf{=}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{ } \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\approx }\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s} \end{gathered}$$

At the instantaneous velocity is, 

$$\begin{gathered} \mathbf{v}\mathbf{=}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{\pi t}}{\mathbf{2}}\mathbf{\right)} \\ \mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s} \end{gathered}$$

At ${}^{\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{s}}$ the instantaneous velocity is, 

Therefore, the instantaneous velocity of the particle at ${}^{\mathbf{t}\mathbf{ }\mathbf{=}\mathbf{o}\mathbf{,}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}\mathbf{ }\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{ }\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$ is ${}^{\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$, ${}^{\mathbf{1}\mathbf{.}\mathbf{1}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$ and ${}^{\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}\mathbf{s}}$ respectively.

Step 5: (c) Determination of the average acceleration between to

Using equation (iii), the average acceleration is calculated as follows: 

$$\begin{gathered} {\mathbf{a}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{v}\left(2.0 s\right)\mathbf{-}\mathbf{v}\left(0 s\right)}{\mathbf{\Delta t}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{57}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{s}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$ 

Thus, the average acceleration of the particle between${}^{\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{0}}$ and ${}^{\mathbf{t}\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$ is${}^{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{ }\mathbf{c}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$

Step 6: (d) Determination of the instantaneous acceleration

Using equation (iv), the instantaneous acceleration is,

$$\begin{gathered} \mathit{a}\mathbf{=}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left\{\left(\frac{\pi }{2}cm\right)sin\left(\frac{\pi }{4}\right)\right\} \\ \mathit{a}\mathbf{=}\mathbf{-}\left(\frac{{\pi }^2}{8}cm\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi }{4}\right) \end{gathered}$$ 

At t = 0 s, the instantaneous acceleration is, 

 $$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{-}\mathbf{\left(}\frac{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{8}}\mathbf{cm}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{0}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$

At t = 1 s, the instantaneous acceleration is, 

 $$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{-}\mathbf{\left(}\frac{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{8}}\mathbf{cm}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{87}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$

At t = 2 s, the instantaneous acceleration is, 

$$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{-}\mathbf{\left(}\frac{{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{8}}\mathbf{cm}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{\pi }}{2}\mathbf{\right)} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{-}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}} \end{gathered}$$

Thus, the instantaneous acceleration of the particle at ${}^{\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{ }}$, ${}^{\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$ and ${}^{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{s}}$ is ${}^{\mathbf{0}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{87}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$ and ${}^{\mathbf{-}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}}$respectively.