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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

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.

A 15 kg rock is dropped from rest on the earth and reaches the ground in \(1.75 \mathrm{~s}\). When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in \(18.6 \mathrm{~s}\). What is the acceleration due to gravity on Enceladus?

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in \(1.90 \mathrm{~s}\). You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick's velocity just before it reaches the ground? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the brick.

A rocket carrying a satellite is accelerating straight up from the earth's surface. At \(1.15 \mathrm{~s}\) after liftoff, the rocket clears the top of its launch platform, \(63 \mathrm{~m}\) above the ground. After an additional \(4.75 \mathrm{~s},\) it is \(1.00 \mathrm{~km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 s part of its flight and (b) the first \(5.90 \mathrm{~s}\) of its flight.

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point \(30.0 \mathrm{~m}\) below its starting point \(5.00 \mathrm{~s}\) after it leaves the thrower's hand. Ignore air resistance. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.
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