91影视

The given data can be listed below as,

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

Newton鈥檚 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.

The given equation can be expressed as:

$v_x\left(t\right)=a-尾t^2$ 鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�� i)

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

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

Substituting the equation i), we get-

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

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

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

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:

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

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

$鈭�-尾t^2=0$

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

Using$t=\sqrt{\frac{鈭�}{尾}}$, the maximum positive displacement of the object is:

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

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鈥檚 maximum positive displacement from the origin is 3.7704 m.