Q68P - University Physics with Modern Physics

Question

An object’s velocity is measured to be ${\mathbf{V}}_{\mathbf{x}}\left(t\right)\mathbf{=}\mathbf{a}\mathbf{-}{\mathbf{\beta t}}^{\mathbf{2}}$, where $\mathbf{a}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$ and $\mathbf{\beta }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}$. At t = 0 the object is at $\mathbf{x}\mathbf{=}\mathbf{0}$ . (a) Calculate the object’s position and acceleration as functions of time. (b) What is the object’s maximum positive displacement from the origin?

Step-by-Step Solution

Answer

a) The position and acceleration of the object as a function of time are $4\mathrm{t}-\frac{2{\mathrm{t}}^3}{3}$and -4t respectively.

b) The object’s maximum positive displacement from the origin is $3.7704 \mathrm{m}.$

Step 1: Identification of the given data

The given data can be listed below as,

• The value of $\propto \mathrm{is} 4.00\mathrm{m}/\mathrm{s}$
• The value of $\mathrm{\beta } \mathrm{is} 2.00\mathrm{m}/{\mathrm{s}}^3$
• The object is at a position x=0.

Step 2: Significance of Newton’s first law in calculating the acceleration, position and maximum positive displacement

Newton’s first law elucidates that one of the main reason for the change in the direction of an object is due to the act of an external force.

 

Differentiating the equation of velocity gives the acceleration and integrating the equation of velocity gives the position. Moreover, the equation of position gives the maximum positive displacement.

Step 3: Determination of position and acceleration and the maximum positive displacement

The given equation can be expressed as:

$v_x\left(t\right)=a-\beta t^2$   ……………………………………………………………………………… i)

a) From Newton’s first law, the position of an object is described as:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{\mathrm{dx}}{\mathrm{dt}} \\ \mathrm{dx}={\mathrm{v}}_{\mathrm{x}}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{x}=\int {\mathrm{v}}_{\mathrm{x}}\left(\mathrm{t}\right)\mathrm{dt} \end{gathered}$$

Substituting the equation i), we get-

$$\begin{gathered} x={\int }_0^t\left(a-\beta t^2\right)dt \\ x={\left[at-\frac{\beta t^3}{3}\right]}_0^t \\ x=at-\frac{\beta t^3}{3} \\ x=4t-\frac{2t^3}{3} \end{gathered}$$

From Newton’s first law, the acceleration of an object is described as:

$$\begin{gathered} {\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{{\mathrm{dv}}_{\mathrm{x}}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ {\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}-{\mathrm{\beta t}}^2\right) \\ {\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=-2\mathrm{\beta t} \\ {\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=-2\times 2\times \mathrm{t} \\ {\mathrm{a}}_{\mathrm{x}}\left(\mathrm{t}\right)=-4\mathrm{t} \end{gathered}$$

Thus, the position and acceleration as a function of time are $4\mathrm{t}-\frac{2{\mathrm{t}}^3}{3} \mathrm{and} -4\mathrm{t}$ respectively.

b) 

Analysing the above solution, the equation of displacement can be expressed as:

$$\begin{gathered} \mathrm{x} \mathrm{or} \mathrm{f}\left(\mathrm{t}\right)=\propto \mathrm{t}-\frac{{\mathrm{\beta t}}^3}{3} \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{t}\right)=\propto -{\mathrm{\beta t}}^2 \end{gathered}$$

As t=0 , the above equation can be expressed as:

$\propto -\beta t^2=0$

Solving the above equation, the value of t becomes $\sqrt{\frac{\propto }{\beta }}$as the other value which is 0 is not considered.

Using $t=\sqrt{\frac{\propto }{\beta }}$, the maximum positive displacement of the object is:

${\mathrm{x}}_{\max }=\propto \sqrt{\frac{\propto }{\mathrm{\beta }}}-\frac{\mathrm{\beta }}{3}{\left(\frac{\propto }{\mathrm{\beta }}\right)}^{3/2}$

Substituting the values of in the above equation, we get-

$$\begin{gathered} {\mathrm{x}}_{\max }=4.00\mathrm{m}/\mathrm{s}\sqrt{\frac{4.00\mathrm{m}/\mathrm{s}}{2.00\mathrm{m}/\mathrm{s}}}-\frac{2.00\mathrm{m}/{\mathrm{s}}^3}{3}{\left(\frac{4.00\mathrm{m}/\mathrm{s}}{2.00\mathrm{m}/{\mathrm{s}}^3}\right)}^{3/2} \\ {\mathrm{x}}_{\max }=5.656\mathrm{m}-1.8856\mathrm{m} \\ {\mathrm{x}}_{\max }=3.7704 \mathrm{m} \end{gathered}$$

Thus, the object’s maximum positive displacement from the origin is 3.7704 m.