Q8E - University Physics with Modern Physics

Question

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

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

(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 $\mathit{t}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$ ?

Step-by-Step Solution

Answer

1. The x and y-components of the car’s velocity as functions of time are $-0.036t \mathrm{m}/{\mathrm{s}}^3$ and $0.55 \mathrm{m}/{\mathrm{s}}^2$ respectively.
2. The magnitude of the velocity of the car is 7.47 m/s and the direction of the velocity of the car is $59°$.
3. 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 $297.6°$

Step 1: Identification of given data

The given data can be listed below,

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

Step 2: Concept/Significance of the acceleration

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.

Step 3: (a) Determination of a x ( t ) and a y ( t ) , the x- and y-components of the car’s velocity as functions of 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,

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

 

The y-component of the acceleration is given by,

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

 

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

Step 4: (b) Determination of the magnitude and direction of the car’s velocity at

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 $59°$.

Step 5: (c) Determination of the magnitude and direction of the car’s acceleration at t = 8.0 s

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 $297.6°$