Question

CALC A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by \(\hat{v} =[5.00 m/s - (0.0180 m/s^3)t^2] \hat{\imath} = [2.00 m/s + (0.550 m/s^2)t]\hat{j}\). (a) What are \(a_x(t)\) and \(a_y(t)\), 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 \(t\) = 8.00 s? (b) What are the magnitude and direction of the car's acceleration at \(t\) = 8.00 s?

Short answer

(a) \( a_x(t) = -0.0360t \), \( a_y(t) = 0.550 \). (b) \( v = 6.98 \text{ m/s}, \) \( 66.8^\circ \). \( a = 0.620 \text{ m/s}^2 \), \( 117^\circ \).

Step-by-step solution

Identify the Velocity Components

From the given velocity vector \( \hat{v} \), we can identify the components:- The velocity in the \( x \)-direction: \( v_x = 5.00 \text{ m/s} - (0.0180 \text{ m/s}^3)t^2 \).- The velocity in the \( y \)-direction: \( v_y = 2.00 \text{ m/s} + (0.550 \text{ m/s}^2)t \).

Find the Acceleration Components

To find the acceleration components, differentiate the velocity components with respect to time:- For the \( x \)-component: \( a_x(t) = \frac{d}{dt} \left(5.00 - 0.0180t^2 \right) = -0.0360t \text{ m/s}^2 \).- For the \( y \)-component: \( a_y(t) = \frac{d}{dt} \left(2.00 + 0.550t \right) = 0.550 \text{ m/s}^2 \).

Evaluate the Velocity Components at t = 8.00 s

Substitute \( t = 8.00 \) s into the expressions for \( v_x \) and \( v_y \):- \( v_x(8) = 5.00 - (0.0180)(8)^2 = 2.85 \text{ m/s} \).- \( v_y(8) = 2.00 + (0.550)(8) = 6.40 \text{ m/s} \).

Calculate the Magnitude and Direction of Velocity

The magnitude of the velocity vector is given by \( v = \sqrt{v_x^2 + v_y^2} \):- \( v = \sqrt{(2.85)^2 + (6.40)^2} = 6.98 \text{ m/s} \).The direction \( \theta \) is the angle with respect to the \( x \)-axis:- \( \theta = \tan^{-1}\left( \frac{v_y}{v_x} \right) = \tan^{-1}\left( \frac{6.40}{2.85} \right) = 66.8^\circ \).

Evaluate the Acceleration Components at t = 8.00 s

Substitute \( t = 8.00 \) s into the expressions for \( a_x \) and \( a_y \):- \( a_x(8) = -0.0360 \times 8 = -0.288 \text{ m/s}^2 \).- \( a_y(8) = 0.550 \text{ m/s}^2 \).

Calculate the Magnitude and Direction of Acceleration

The magnitude of the acceleration vector is given by \( a = \sqrt{a_x^2 + a_y^2} \):- \( a = \sqrt{(-0.288)^2 + (0.550)^2} = 0.620 \text{ m/s}^2 \).The direction \( \phi \) is the angle with respect to the \( x \)-axis:- \( \phi = \tan^{-1}\left( \frac{a_y}{a_x} \right) = \tan^{-1}\left( \frac{0.550}{-0.288} \right) = 117^\circ \), meaning it is directed above the negative \( x \)-axis.

Key concepts

Velocity Components

Understanding the velocity components of an object is crucial in kinematics, as they help us determine how fast something is moving in a particular direction. In the case of the remote-controlled car, the velocity is given as a vector function of time:
\[\hat{v} =[5.00 \text{ m/s} - (0.0180 \text{ m/s}^3)t^2] \hat{\imath} + [2.00 \text{ m/s} + (0.550 \text{ m/s}^2)t]\hat{j}\]
Here are the components:

• The velocity in the x-direction: \( v_x = 5.00 \text{ m/s} - (0.0180 \text{ m/s}^3)t^2 \)
• The velocity in the y-direction: \( v_y = 2.00 \text{ m/s} + (0.550 \text{ m/s}^2)t \)

These equations tell us how the car's speed changes over time in the horizontal and vertical directions. At any time \( t \), you can substitute into these equations to find the speed of the car in the x and y directions. This is a foundational concept for solving problems in kinematics, as it allows for the decomposition of complex motion into simpler, perpendicular directional components.

Acceleration Components

The concept of acceleration is all about how the velocity of an object changes over time. In the context of the given exercise, the acceleration components can be found by differentiating the velocity components with respect to time.
For our remote-controlled car:

• The acceleration in the x-direction is obtained by differentiating \( v_x \) with respect to time: \( a_x(t) = \frac{d}{dt}[5.00 - 0.0180t^2] = -0.0360t \text{ m/s}^2 \).
• The acceleration in the y-direction is constant, since we differentiate \( v_y \): \( a_y(t) = \frac{d}{dt}[2.00 + 0.550t] = 0.550 \text{ m/s}^2 \).

Knowing these components, you can describe how the velocity is changing in both the horizontal and vertical directions. If you substitute a specific time into the equations for \( a_x \) and \( a_y \), you will discover how the object's velocity is accelerating at that moment. This is incredibly useful for predicting future motion and making real-time adjustments. Think of it as understanding the push and pull in each axis independently.

Magnitude and Direction

Magnitude and direction are essential when dealing with vector quantities like velocity and acceleration. They help us convert the vector into an understandable single value and its orientation in space.
To find the magnitude of velocity or acceleration, we use the formula:

• For velocity: \( v = \sqrt{v_x^2 + v_y^2} \)
• For acceleration: \( a = \sqrt{a_x^2 + a_y^2} \)

The direction is determined by the angle formed with the positive x-axis. For velocity, this is given by:

• \( \theta = \tan^{-1}\left( \frac{v_y}{v_x} \right) \)

And for acceleration:

• \( \phi = \tan^{-1}\left( \frac{a_y}{a_x} \right) \)

Let's apply this at \( t = 8.00 \text{ s} \):

• Velocity magnitude: \( v = \sqrt{(2.85)^2 + (6.40)^2} = 6.98 \text{ m/s} \)
• Direction with respect to the x-axis: \( \theta = 66.8^\circ \)
• Acceleration magnitude: \( a = \sqrt{(-0.288)^2 + (0.550)^2} = 0.620 \text{ m/s}^2 \)
• Direction with respect to the x-axis: \( \phi = 117^\circ \)

Thus, magnitude gives us a scalar context while direction provides the spatial insight. Together, they let us fully appreciate the behavior of the moving car in the parking lot. By mastering these concepts, you develop a deeper understanding of how vectors shape motion equations in physics.