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Chapter 3: Problem 8

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

Short Answer

Expert verified
The x-component of the acceleration \(a_{x}(t)\) is \(-0.036 \mathrm{m/s^2} t\), and the y-component \(a_{y}(t)\) is \(0.550 \mathrm{m/s^2}\). The magnitude of the velocity at \(t=8.00 \mathrm{s}\) is \(6.40 \mathrm{m/s}\) with a direction of \(19.4^\circ\) north of east. The magnitude of acceleration is \(0.735 \mathrm{m/s^2}\) with a direction of \(56.6^\circ\) south of east.

Step by step solution

01

Find \(a_x(t)\) and \(a_y(t)\)

These are the x- and y-components of the car's velocity as functions of time. As acceleration is the derivative of velocity with respect to time, simply differentiate the function given for the x and y components of the velocity vector as follows:\n\nTo find \(a_x(t)\), differentiate the x-component of the velocity: \n\n\[a_x(t) = \frac{d}{dt} [5.00 \mathrm{~m/s} - (0.0180 \mathrm{~m/s^3}) t^2]\] \n\nSimilarly, for \(a_y(t)\), differentiate the y-component of the velocity: \n\n\[a_y(t) = \frac{d}{dt} [2.00 \mathrm{m/s} + (0.550 \mathrm{m/s^2}) t]\]
02

Calculate Magnitude and Direction of the Car's Velocity at \(t = 8.00 \mathrm{s}\)

To find the magnitude of the velocity, substitute \(t = 8.00 \mathrm{s}\) into the equation of the velocity vector and find the magnitude with: \n\n\[\| \overrightarrow{v} \| = \sqrt{v_x^2 + v_y^2}\] \n\nthen, find the direction, \(\theta\), with: \n\n\[\theta = \arctan \left( \frac{v_y}{v_x} \right)\] which is the angle made with the positive x-axis. Use inverse tangent function to find the angle and ensure it lies within the appropriate range.
03

Calculate Magnitude and Direction of the Car's Acceleration at \(t = 8.00 \mathrm{s}\)

Similar to Step 2, first find \(a_x(8.00 \mathrm{s})\) and \(a_y(8.00 \mathrm{s})\) by substituting \(t = 8.00 \mathrm{s}\) into equations obtained in Step 1 for x and y components of acceleration. The magnitude of the acceleration vector can be found out by using the formula \n\n\[\| \overrightarrow{a} \| = \sqrt{a_x^2 + a_y^2}\] \n\nThen, the direction, \(\phi\), of the acceleration is given by \n\[\phi = \arctan \left( \frac{a_y}{a_x} \right)\] where, as before, \(\phi\) is the angle made with the positive x-axis. Use the inverse tangent function to find the angle, again ensuring it is within the appropriate range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Velocity and Acceleration
When we speak of velocity and acceleration in physics, we are often referring to vector quantities. A vector quantity is one that has both magnitude (how much) and direction (which way). For example, if a car moves eastward at 60 kilometers per hour, its velocity vector includes both the speed (the magnitude, 60 km/h) and the direction (eastward).

Acceleration is also a vector quantity, representing how the velocity of an object changes with time. If an object speeds up, slows down, or changes direction, it is experiencing acceleration. The process of finding the acceleration vector involves calculating the rate of change of the velocity vector over time. This is known as time-dependent acceleration, which refers to how the acceleration changes as time progresses.

Just like velocity, the acceleration vector has an x-component and a y-component, especially when dealing with two-dimensional motion. These components describe how the velocity is changing in each direction of a coordinate system. It's crucial to understand the vector nature of these quantities to fully comprehend the motion of objects.
Time-Dependent Acceleration
Time-dependent acceleration is the phenomenon where an object's acceleration changes as time goes on. This can happen in scenarios where forces are applied in different magnitudes during the motion or when the force direction changes with time. In the example of the remote-controlled car, the acceleration in the x-direction is affected by a time-squared term, and in the y-direction, it's affected by a time term. This shows that the car's acceleration components are not constant; they depend on t, the time variable.

The x-component of the acceleration, or a_x(t), is found by differentiating the x-component of the velocity vector with respect to time. Likewise, the y-component of the acceleration, or a_y(t), is found by differentiating the y-component of the velocity vector with respect to time. Understanding these concepts is fundamental for solving problems involving motion where acceleration is not constant.
Magnitude and Direction of Velocity
The magnitude of velocity is equivalent to the speed of an object but taken with a consideration for direction, making it a vector. To determine the magnitude of the car's velocity at any given moment, we use the Pythagorean theorem on the velocity components. Specifically, it involves squaring each component, adding them together, and then taking the square root of the sum.

Moreover, the direction of velocity is equally important as it tells us where the object is heading. The direction of the car's velocity vector at any point in time can be found by calculating the arctangent of the ratio of the y-component to the x-component of the velocity. This angle is measured from the positive x-axis and is crucial for visualizing the object's trajectory. Understanding both magnitude and direction provides a comprehensive picture of the car's motion.
Magnitude and Direction of Acceleration
Similar to velocity, the acceleration of an object can be described in terms of its magnitude and direction. The magnitude of acceleration tells us how quickly the velocity of the object is changing, while the direction reveals the line along which this change in velocity is happening. To calculate the magnitude of the acceleration, we use the method similar to finding the magnitude of velocity: square both the x and y components of acceleration, add them up, and take the square root of the total.

To find the direction of acceleration, we also use the arctangent function, but this time it's for the ratio of the y-component to the x-component of the acceleration vector. The result will give us an angle that conveys the direction of acceleration with respect to the positive x-axis. Comprehending the acceleration's magnitude and direction is vital since it helps predict how an object's motion will change over time.

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Most popular questions from this chapter

A jet plane is flying at a constant altitude. At time \(t_{1}=0,\) it has components of velocity \(v_{x}=90 \mathrm{~m} / \mathrm{s}, v_{y}=110 \mathrm{~m} / \mathrm{s} .\) At time \(t_{2}=30.0 \mathrm{~s}\) the components are \(v_{x}=-170 \mathrm{~m} / \mathrm{s}, v_{y}=40 \mathrm{~m} / \mathrm{s}\). (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

A small ball is attached to the lower end of a 0.800 -m-long string, and the other end of the string is tied to a horizontal rod. The string makes a constant angle of \(37.0^{\circ}\) with the vertical as the ball moves at a constant speed in a horizontal circle. If it takes the ball \(0.600 \mathrm{~s}\) to complete one revolution, what is the magnitude of the radial acceleration of the ball?

A faulty model rocket moves in the \(x y\) -plane (the positive \(y\) -direction is vertically upward). The rocket's acceleration has components \(a_{x}(t)=\alpha t^{2}\) and \(a_{y}(t)=\beta-\gamma t,\) where \(\alpha=2.50 \mathrm{~m} / \mathrm{s}^{4}\) \(\beta=9.00 \mathrm{~m} / \mathrm{s}^{2},\) and \(\gamma=1.40 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the rocket is at the origin and has velocity \(\overrightarrow{\boldsymbol{v}}_{0}=v_{0 x} \hat{\imath}+v_{0 y} \hat{\jmath}\) with \(v_{0 x}=1.00 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=7.00 \mathrm{~m} / \mathrm{s} .\) (a) Calculate the velocity and position vectors as functions of time. (b) What is the maximum height reached by the rocket? (c) What is the horizontal displacement of the rocket when it returns to \(y=0 ?\)

A river flows due south with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). You steer a motorboat across the river; your velocity relative to the water is \(4.2 \mathrm{~m} / \mathrm{s}\) due east. The river is \(500 \mathrm{~m}\) wide. (a) What is your velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of your starting point will you reach the opposite bank?

The nose of an ultralight plane is pointed due south, and its airspeed indicator shows \(35 \mathrm{~m} / \mathrm{s}\). The plane is in a \(10 \mathrm{~m} / \mathrm{s}\) wind blowing toward the southwest relative to the earth. (a) In a vectoraddition diagram, show the relationship of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\) (the velocity of the plane relative to the earth) to the two given vectors. (b) Let \(x\) be east and \(y\) be north, and find the components of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\). (c) Find the magnitude and direction of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\)
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