An object is moving along the x-axis. At t = 0 it has velocity ${\mathbf{v}}_{\mathbf{0}}\mathbf{x}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$. Starting at time t = 0 it has acceleration ${\mathbf{a}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{Ct}$, where C has units of $\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{3}}$. (a) What is the value of C if the object stops in 8.00 s after t = 0? (b) For the value of C calculated in part (a), how far does the object travel during the 8.00 s?

The given data can be listed below as,

• The velocity of the object at time t = 0 is, ${\mathrm{v}}_{0\mathrm{x}}=20.0\mathrm{m}/\mathrm{s}$
• The expression for the acceleration at time t = 0 is,$a_x=-Ct$

It is known that the velocity with respect to time is known as acceleration. So, acceleration can be obtained by taking the derivative of the function of velocity with respect to time. It is expressed as follows,

$\mathit{a}\mathbf{=}\frac{\mathbf{d}\mathbf{v}}{\mathbf{dt}}$

Here, v is the velocity of the object and t is the time period.

Also, the acceleration can be obtained by taking the double derivative of the function of position with respect to time. It is expressed as follows,

$a=\frac{d^{2}x}{d{t}^2}$

Here, x is the position of the object.

(a)

Write the expression for the integration of dv in the limits$v_{0_x}$ and $v_x$.

${\int }_{v_{0_x}}^{v_x}dv={\int }_0^t{a}_{x}dt$

Substitute all the values in the above expression.

$$\begin{gathered} {\int }_{v_{0x}}^{v_x} dv={\int }_0^8 -Ctdt \\ [v{]}_{20}^0=-C{\left[\frac{t^2}{2}\right]}_0^8 \\ [0-20]=-C\left[\frac{8^2}{2}-0\right] \\ -20=-C\left(32\right) \\ C=\frac{20}{32} \\ =0.625\mathrm{m}/{\mathrm{s}}^3 \end{gathered}$$

Thus, the value of C, if the object stops after 8 s is $0.625m/s^3$.

(b)

Write the expression for the integration of the position.

$$\begin{gathered} a_x=\frac{dt}{dt}=\frac{d^{2}x}{d{t}^2} \\ -Ct=\frac{d^{2}x}{d{t}^2} \\ \frac{d^{2}x}{d{t}^2}=-Ct \end{gathered}$$

Integrate the above expression.

$$\begin{gathered} \frac{dx}{dt}=-C\frac{t^2}{2}+D \\ v=-C\frac{t^2}{2}+D \end{gathered}$$ …(1)

Substitute all the values in the above expression at time t = 0.

$$\begin{gathered} 20=0+D \\ D=20m/s \end{gathered}$$

Integrate the expression$\frac{dx}{dt}=-C\frac{t^2}{2}+D\mathrm{w}$ within the required limits.

$$\begin{gathered} {\int }_0^x dx={\int }_0^8 \left(-C\frac{t^2}{2}+D\right)dt \\ [x{]}_0^x={\left[-\frac{C}{2}\frac{t^3}{3}+Dt\right]}_0^8 \\ (x-0)=\left(-\frac{C}{6}{8}^3+D\left(8\right)+0-0\right) \\ x=-\frac{0.625}{6}{8}^3+\left(20\right)\left(8\right) \\ =213.334\mathrm{m} \end{gathered}$$

Thus, the position of the object after the time period of 8 s is 213.334 m.