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Chapter 2: Problem 68

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\), where \(\alpha=4.00 \mathrm{~m} / \mathrm{s}\) and \(\beta=2.00 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0 .\) (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum positive displacement from the origin?

Short Answer

Expert verified
a) The position as a function of time is \(x(t) = \alpha t - \frac{\beta t^{3}}{3}\). The acceleration as a function of time is \(a(t) = -2\beta t\). b) The maximum positive displacement from the origin is \(x_{max} = \alpha \sqrt{\frac{\alpha}{\beta}} - \frac{\beta}{3} \left(\sqrt{\frac{\alpha}{\beta}}\right)^3\).

Step by step solution

01

The Acceleration Function

First, the acceleration function can be calculated as the second derivative of the velocity function with respect to time. That means we take the derivative of \(v_{x}(t)=\alpha-\beta t^{2}\) once. Since \(\alpha\) is a constant, its derivative will be 0. As for \(-\beta t^{2}\), applying the power rule for derivatives (for any real number n different from 0, the derivative of \(t^n\) is \(nt^{n鈭�1}\)), we have \(2*\beta t\). Hence, the acceleration as a function of time is \(a(t) = -2\beta t\).
02

The Position Function

Integrating the velocity function gives us the position as a function of time. The integral \(\int v(t) dt\) yields \(\int (\alpha-\beta t^{2}) dt\). Performing the integral operation we get \(\alpha t - \frac{\beta t^{3}}{3}+c\), however since at \(t=0\) the object was at position \(x=0\) \(c = 0\). Thus, \(x(t) = \alpha t - \frac{\beta t^{3}}{3} \).
03

Maximum Positive Displacement

The maximum positive displacement is obtained when the velocity equals 0, because at that point the object stops moving to the right (positive displacement) and starts moving to the left (negative displacement). Setting the velocity equal to zero gives us \(\alpha-\beta t^2 = 0\). Solving for \(t\) yields \(t = \sqrt{\frac{\alpha}{\beta}}\). Substituting \(t = \sqrt{\frac{\alpha}{\beta}}\) into the \(x(t)\) yields the maximum displacement, \(x_{max} = \alpha \sqrt{\frac{\alpha}{\beta}} - \frac{\beta}{3} \left(\sqrt{\frac{\alpha}{\beta}}\right)^3\).

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Key Concepts

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

Velocity Function
In kinematics, the velocity function describes how an object's velocity changes over time. For the exercise in question, the object's velocity function is given by \(v_{x}(t)=\alpha-\beta t^{2}\). This means the velocity is dependent on two constants: \(\alpha\) and \(\beta\), which are given as 4.00 m/s and 2.00 m/s鲁 respectively. The formula highlights that the velocity starts from a constant value and decreases as time progresses due to the presence of the \(t^2\) term in the equation.
  • When \(t = 0\), the velocity is \(v_{x}(0) = \alpha = 4.00 \mathrm{~m} / \mathrm{s}\).
  • As \(t\) increases, the term \(-\beta t^2\) becomes more significant, decreasing the velocity.
  • The velocity function is integral in determining other aspects such as displacement and when the object changes direction.
Acceleration Function
Acceleration is the rate of change of velocity with respect to time. To find the acceleration function for a given velocity function, we take the derivative of the velocity function with respect to time. Applying this to \(v_{x}(t)=\alpha-\beta t^{2}\), we differentiate each term: - The derivative of the constant \(\alpha\) is 0. - Using the power rule, the derivative of \(-\beta t^2\) is \(-2\beta t\). So, the acceleration function becomes \(a(t) = -2\beta t\).This tells us:
  • The acceleration is directly proportional to time \(t\), meaning it changes as time passes.
  • With \(\beta\) being constant, the only variable affecting acceleration's magnitude and direction over time is \(t\).
  • The negative sign indicates that the object is decelerating over time, slowing down as \(t\) increases.
Displacement Calculation
Displacement is the overall change in position of the object. To calculate the displacement, we must integrate the velocity function over time. From the velocity function \(v_{x}(t)=\alpha-\beta t^{2}\), integrating gives \(x(t) = \alpha t - \frac{\beta t^3}{3} + c\). Using the initial condition that at \(t=0\), \(x=0\), we find the constant \(c = 0\). So, the position function simplifies to \(x(t) = \alpha t - \frac{\beta t^3}{3}\).This position (or displacement) function tells us:
  • The object's displacement is initially determined by the \(\alpha t\) term.
  • The cubic term \(-\frac{\beta t^3}{3}\) modifies the displacement depending on time, showing more complexity than a linear motion.
  • It accurately traces the object's path over time, indicating how the initial motion is progressively counteracted by deceleration.
Integration in Kinematics
Integration in kinematics is a crucial tool for connecting velocity and position. It allows us to determine an object's displacement from its velocity function by performing an integral. Understanding integration helps us gain insights into motion behaviors like distance traveled and changes in direction. When we have a velocity function like \(v_{x}(t)=\alpha-\beta t^{2}\), integrating it over time gives us the position function \(x(t)\). The integration accounts for all the changes in velocity over the given time, summing the small displacements to provide a complete picture of motion.Key integration points in kinematics include:
  • The antiderivative of a function returns the position function when integrating the velocity function.
  • We must always consider initial conditions to solve for constants introduced through integration (e.g., \(c\) in the displacement formula).
  • Integration is not only pivotal for displacement but also for calculating area under velocity vs. time graphs, representing total distance.

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Most popular questions from this chapter

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of \(20 \mathrm{~m} / \mathrm{s}\) (45 mi/h) when it reaches the end of the 120 -m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of \(20 \mathrm{~m} / \mathrm{s}\). What distance does the traffic travel while the car is moving the length of the ramp?

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled \(6.80 \mathrm{~m}\) to the bottom of the incline is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline?

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of \(960 \mathrm{~m}\) above the earth's surface. The rocket's engines give the rocket an upward acceleration of \(16.0 \mathrm{~m} / \mathrm{s}^{2}\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.
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