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The acceleration of a motorcycle is given by \(a_{x}(t)=\) \(A t-B t^{2},\) where \(A=1.50 \mathrm{m} / \mathrm{s}^{3}\) and \(B=0.120 \mathrm{m} / \mathrm{s}^{4} .\) The motorcycle is at rest at the origin at time \(t=0\) . (a) Find its position and velocity as functions of time. (b) Calculate the maximum velocity it attains.

Short Answer

Expert verified
Position function: \(x(t) = \frac{1}{6}At^3 - \frac{1}{12}Bt^4\). Velocity function: \(v_x(t) = \frac{1}{2}At^2 - \frac{1}{3}Bt^3\). Maximum velocity: 46.87 m/s.

Step by step solution

01

Understand the Given Information

We know that the acceleration of the motorcycle is expressed as a function of time: \(a_x(t) = At - Bt^2\), with the constants \(A = 1.50 \text{ m/s}^3\) and \(B = 0.120 \text{ m/s}^4\). The motorcycle starts at rest from the origin at \(t = 0\). Our task is to find the velocity and position functions, and to determine the maximum velocity.
02

Derive the Velocity Function

Velocity is the integral of acceleration. Integrate \(a_x(t) = At - Bt^2\) with respect to time to get the velocity function \(v_x(t)\). This gives:\[v_x(t) = \int (At - Bt^2) \, dt = \frac{1}{2}At^2 - \frac{1}{3}Bt^3 + C\]Since the motorcycle is initially at rest, \(v_x(0) = 0\), we solve for \(C\):\[0 = \frac{1}{2}A(0)^2 - \frac{1}{3}B(0)^3 + C\]Thus, \(C = 0\) and the velocity function is:\[v_x(t) = \frac{1}{2}At^2 - \frac{1}{3}Bt^3\].
03

Derive the Position Function

Position is the integral of velocity. Integrate the velocity function \(v_x(t) = \frac{1}{2}At^2 - \frac{1}{3}Bt^3\):\[x(t) = \int \left(\frac{1}{2}At^2 - \frac{1}{3}Bt^3\right) \, dt = \frac{1}{6}At^3 - \frac{1}{12}Bt^4 + D\]Initially, the position is zero \(x(0) = 0\), so we solve for \(D\):\[0 = \frac{1}{6}A(0)^3 - \frac{1}{12}B(0)^4 + D\]Thus, \(D = 0\) and the position function is:\[x(t) = \frac{1}{6}At^3 - \frac{1}{12}Bt^4\].
04

Find the Time When Velocity is Maximum

The maximum velocity occurs when the acceleration is zero (change in acceleration from positive to negative), meaning \(\frac{dv_x}{dt} = 0\). So, set the given acceleration \(a_x(t) = At - Bt^2\) to zero:\[At - Bt^2 = 0\]Solving for \(t\), we have:\[t(At - Bt) = 0\]\[t(A - Bt) = 0\]\(t = 0\) or \(t = \frac{A}{B}\).Since \(t=0\) is when the motion starts, the relevant time for maximum velocity is \(t = \frac{A}{B}\).
05

Calculate the Maximum Velocity

Substitute \(t = \frac{A}{B}\) into the velocity function \(v_x(t)\):\[v_{max} = \frac{1}{2}A\left(\frac{A}{B}\right)^2 - \frac{1}{3}B\left(\frac{A}{B}\right)^3\]Simplify:\[v_{max} = \frac{1}{2} \frac{A^3}{B^2} - \frac{1}{3} \frac{A^3}{B^2}\]\[v_{max} = \left(\frac{1}{2} - \frac{1}{3}\right) \frac{A^3}{B^2} = \frac{1}{6} \frac{A^3}{B^2}\].Thus, calculate the numerical value:\[v_{max} = \frac{1}{6} \frac{(1.50)^3}{(0.120)^2}\].After calculation, the maximum velocity is approximately 46.87 m/s.

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Key Concepts

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

Kinematics equations
Kinematics equations are essential tools in physics used to describe the motion of objects. These equations consider aspects like velocity, acceleration, and position, allowing us to predict future states of a moving object. In this exercise, the acceleration of the motorcycle is provided as a function:
  • \( a_x(t) = A t - B t^2 \)
with the constants \(A = 1.50 \, \mathrm{m/s}^3\) and \(B = 0.120 \, \mathrm{m/s}^4\).
Knowing this form, we can derive other kinematic quantities such as velocity and displacement by integrating the acceleration function.
To solve for the velocity, one must understand that it’s essentially an antiderivative of acceleration. Additionally, the initial conditions—like starting from rest at the origin—are crucial for solving the equations.
Integration in physics
Integration is a mathematical process used to find quantities like velocity and position as functions of time. It is the reverse process of differentiation and plays a central role in physics problems involving changing quantities.
When given an acceleration function, say \( a_x(t) = At - Bt^2 \), integrating this with respect to time provides the velocity function:
  • \( v_x(t) = \int (At - Bt^2) \, dt = \frac{1}{2}At^2 - \frac{1}{3}Bt^3 + C \)
The constant \(C\) is determined using initial conditions, here \( v_x(0) = 0 \) as given, resulting in \(C = 0\).
Further integration of this velocity function yields the position function. The constant in the position function is similarly found using initial conditions, commonly that the starting position at \(t = 0\) is zero.
Motion analysis
Motion analysis involves understanding how different aspects of motion, such as velocity and displacement, evolve over time. This is a fundamental aspect of physics and takes into account the nature of forces acting on the object.
By analyzing the motion of the motorcycle described in the problem, one must keep track of changes from rest to peak performance. The velocity function derived from acceleration gives insights into the object's speed at any given moment.
Moreover, using the equations of motion allows us to predict the maximum velocity and how quickly the object will travel over certain distances. This deeper examination explains not only how motion begins but also how it changes over shorter periods due to varying amounts of acceleration.
Acceleration function
An acceleration function describes how the speed of an object increases or decreases over time. It often depends on time itself, indicating that acceleration is not constant. In this case, the function is given by:
  • \( a_x(t) = At - Bt^2 \)
The constants \(A\) and \(B\) control how the acceleration changes. These determine the rate of change for velocity and consequently, the motion of the motorcycle.
Finding when this function equals zero is key to understanding motion. It indicates when the object will stop accelerating and reach its maximum velocity.
To anticipate when speed peaks, solve the equation \( At - Bt^2 = 0 \), finding the critical point where acceleration changes direction.
This helps not only in predicting the motion but also offers insight into forces required to move or stop an object, making acceleration functions vital in real-world applications.

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Most popular questions from this chapter

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