A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by

$\overrightarrow{\mathbf{v}}\mathbf{=}\left[5.0m/s-\left(0.0180m/s^3\right)t^2\right]\hat{\mathbf{i}}\mathbf{+}\left(2.0m/s+0.550m/s^{2}t\right)\hat{\mathbf{j}}$.

(a) What are${\mathit{a}}_{\mathbf{x}}\left(t\right)$and${\mathit{a}}_{\mathbf{y}}\left(t\right)$, the x- and y-components of the car’s velocity as functions of time?

(b) What are the magnitude and direction of the car’s velocity at$\mathit{t}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$?

(c) What are the magnitude and direction of the car’s acceleration at$\mathit{t}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$?

The given data can be listed below,

• The velocity of the car is,$\overrightarrow{v}=\left[5.0m/s-\left(0.0180m/s^3\right)t^2\right]\hat{i}+\left(2.0m/s+0.550m/s^{2}t\right)\hat{j}$
• The time at which the velocity of the car is to be found is, $t=8s$.

In the same way that velocity is the derivative of position with respect to time, acceleration is the derivative of velocity with respect to time.

According to the definition, the acceleration is given by,

$a=\frac{dv}{dt}$

Here, v is the velocity of the car.

The x-component of the acceleration is given by,

$$\begin{gathered} a_x\left(t\right)=\frac{d\left(5-0.0180t^2\right)}{dt} \\ =-0.036tm/s^3 \end{gathered}$$

The y-component of the acceleration is given by,

$$\begin{gathered} a_y\left(t\right)=\frac{d\left(2+0.55t\right)}{dt} \\ =0.55m/s^2 \end{gathered}$$

Thus, the x and y-components of the car’s velocity as functions of time are $-0.036t\mathrm{m}/{\mathrm{s}}^2$and $0.55m/s^2$respectively.

The velocity vector for the car at is given by,

$$\begin{gathered} \overrightarrow{v}=\left(5-0.018\times 8^2\right)\hat{i}+\left(2+0.55\times 8\right)\hat{j} \\ =3.848\hat{i}+6.4\hat{j} \end{gathered}$$

The magnitude of the velocity is given by,

$\left|v\right|=\sqrt{v_x^2+v_y^2}$

Substitute all the values in the above expression.

$$\begin{gathered} \left|v\right|=\sqrt{{\left(3.84\mathrm{m}/\mathrm{s}\right)}^2+{\left(6.4\mathrm{m}/\mathrm{s}\right)}^2} \\ =7.47\mathrm{m}/\mathrm{s} \end{gathered}$$

The direction of the velocity of the car is given by,

$$\begin{gathered} \tan \alpha =\frac{v_x}{v_y} \\ \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{6.4}{3.84}\right) \\ =59° \end{gathered}$$

Thus, the magnitude of the velocity of the car is$7.47\mathrm{m}/\mathrm{s}$ and the direction of the velocity of the car is $59°$.

The acceleration vector at the time t = 8 s is given by,

$$\begin{gathered} \overrightarrow{a}=\left(-0.036\times 8\right)\hat{i}+0.55\hat{j} \\ =-0.288\hat{i}+0.55\hat{j} \end{gathered}$$

The magnitude of the acceleration is given by,

$$\begin{gathered} \left|a\right|=\sqrt{{\left(-0.288\mathrm{m}/{\mathrm{s}}^2\right)}^2+{\left(0.55\mathrm{m}/{\mathrm{s}}^2\right)}^2} \\ =0.621\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The direction of the acceleration of the car is given by,

$$\begin{gathered} \tan \alpha =\frac{a_y}{a_x} \\ \alpha ={\tan }^{-1}\left(\frac{a_y}{a_x}\right) \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{0.55}{-0.288}\right) \\ =-62.4° \\ =297.6° \end{gathered}$$

Thus, the magnitude of the acceleration of the car is$0.621\mathrm{m}/{\mathrm{s}}^2$ and the direction of the acceleration of the car is $297.6°$