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Chapter 2: Problem 6

The acceleration of a particle is given by \(a=-k t^{2}\) where \(a\) is in meters per second squared and the time \(t\) is in seconds. If the initial velocity of the particle at \(t=0\) is \(v_{0}=12 \mathrm{m} / \mathrm{s}\) and the particle takes 6 seconds to reverse direction, determine the magnitude and units of the constant \(k .\) What is the net displacement of the particle over the same 6 -second interval of motion?

Short Answer

Expert verified
The constant \(k\) is \(\frac{1}{6} \text{ m/s}^3\) and the net displacement is 54 meters.

Step by step solution

01

Identify Known Quantities

We know that the acceleration of the particle is given by the formula \(a = -k t^2\), the initial velocity \(v_0 = 12 \text{ m/s}\), and the particle reverses direction at \(t = 6 \text{ s}\). Our task is to find the constant \(k\) and the net displacement over the same time interval.
02

Relate Acceleration to Velocity

To find velocity as a function of time, we integrate the acceleration. Starting with \(a = \frac{dv}{dt} = -k t^2\), integrating both sides with respect to \(t\) gives \(v(t) = \int -k t^2 dt = -\frac{k}{3} t^3 + C\). The constant \(C\) will be determined using the initial condition.
03

Incorporate Initial Condition to Find Constant in Velocity Equation

Using the initial condition \(v(0) = v_0 = 12\), we substitute into the velocity equation: \(v(0) = -\frac{k}{3}(0)^3 + C = 12\), so \(C = 12\). Thus, the velocity equation is \(v(t) = -\frac{k}{3} t^3 + 12\).
04

Determine Constant k Using Direction Reversal

When the particle reverses direction, its velocity is zero. Thus, \(v(6) = 0\) at \(t=6\). Substitute \(t = 6\) into \(v(t) = -\frac{k}{3} t^3 + 12 = 0\). This yields \(\frac{k}{3} (6)^3 = 12\), or \(-\frac{k}{3} \times 216 + 12 = 0\). Solving for \(k\) gives \(216k = 36\), so \(k = \frac{36}{216} = \frac{1}{6}\) \(\text{m/s}^3\).
05

Calculate Displacement Using Velocity Function

To find displacement, integrate the velocity function \(v(t) = -\frac{1}{18}t^3 + 12\) from \(t = 0\) to \(t = 6\). The displacement \(s\) is \(\int_0^6 \left(-\frac{1}{18}t^3 + 12\right) dt\), which simplifies to \(-\frac{1}{72} t^4 + 12t\) evaluated from 0 to 6.
06

Evaluate the Displacement Integral

Substitute \(t=6\) into the displacement function: \ \(-\frac{1}{72}(6)^4 + 12(6) = -\frac{1}{72} \times 1296 + 72\).\Calculate \ \(-18 + 72 = 54\).\The net displacement over the 6-second interval is 54 meters.

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Key Concepts

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

Acceleration
Acceleration refers to the rate of change of velocity over time. In this exercise, the acceleration is given as a function of time: \(a = -k t^2\). This expression shows that acceleration changes according to the square of the time and is affected by a constant \(k\). It indicates that as time progresses, the magnitude of acceleration increases, but because of the negative sign, it does so in the negative direction. This decelerates the particle until it eventually comes to a stop and reverses its direction, demonstrating how acceleration impacts the motion of an object over time.
Initial Velocity
Initial velocity is the speed at which a particle starts its motion, in this case, \(v_0 = 12 \text{ m/s}\) at \(t = 0\). Knowing the initial velocity is crucial because it provides a starting point for understanding the overall motion.
When the particle begins its journey, this initial velocity strengthens our understanding of its kinematic behavior. As time progresses, the velocity will adjust according to the integrated acceleration function, allowing us to track its motion comprehensively.
Displacement
Displacement is the change in position of the particle over a period of time. Unlike distance, which is scalar, displacement considers direction and is a vector quantity. In the context of this problem, the net displacement is calculated over a 6-second interval using the velocity function.
By integrating the velocity function from 0 to 6 seconds, we find the total area under the velocity-time graph, representing the displacement. This results in a computed displacement of 54 meters, indicating how far and in which direction the particle travels over the specified timeframe.
Integration
Integration is a mathematical technique used to find quantities like velocity and displacement from acceleration. Since acceleration is the derivative of velocity, integrating it gives us the velocity function. In this exercise, the acceleration \(a = -k t^2\) was integrated to determine the velocity function \(v(t) = -\frac{k}{3}t^3 + C\). To find the net displacement, we then integrate the velocity function over the given time period. This process illustrates how integration helps transition from knowing how the acceleration affects motion to understanding the resulting path the particle takes.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It focuses on parameters such as displacement, velocity, and acceleration. In this problem, kinematics is used to explore how a particle moves when subjected to a time-dependent acceleration.
By understanding the relationships between these quantities through integration, we learn how the particle's motion evolves over time. Kinematics provides the foundation for predicting and analyzing the motion, making it an essential part of understanding particle dynamics.

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Most popular questions from this chapter

A helicopter approaches a rescue scene. A victim \(P\) is drifting along with the river current of speed \(v_{C}=2 \mathrm{m} / \mathrm{s} .\) The wind is blowing at a speed \(v_{W}=\) \(3 \mathrm{m} / \mathrm{s}\) as indicated. Determine the velocity relative to the wind which the helicopter must acquire so that it maintains a steady overhead position relative to the victim.

In a test of vertical leaping ability, a basketball player (a) crouches just before jumping, (b) has given his mass center \(G\) a vertical velocity \(v_{0}\) at the instant his feet leave the surface, and \((c)\) reaches the maximum height. If the player can raise his mass center 3 feet as shown, estimate the initial velocity \(v_{0}\) of his mass center in position ( \(b\) ).

Train \(A\) is traveling at a constant speed \(v_{A}=\) \(35 \mathrm{mi} / \mathrm{hr}\) while car \(B\) travels in a straight line along the road as shown at a constant speed \(v_{B}\). A conductor \(C\) in the train begins to walk to the rear of the train car at a constant speed of \(4 \mathrm{ft} / \mathrm{sec}\) relative to the train. If the conductor perceives car \(B\) to move directly westward at \(16 \mathrm{ft} / \mathrm{sec},\) how fast is the car traveling?

At time \(t=0,\) the 1.8 -lb particle \(P\) is given an initial velocity \(v_{0}=1 \mathrm{ft} / \mathrm{sec}\) at the position \(\theta=0\) and subsequently slides along the circular path of radius \(r=1.5 \mathrm{ft}\). Because of the viscous fluid and the effect of gravitational acceleration, the tangential acceleration is \(a_{t}=g \cos \theta-\frac{k}{m} v,\) where the constant \(k=0.2 \mathrm{lb}\) -sec/ft is a drag parameter. Determine and plot both \(\theta\) and \(\dot{\theta}\) as functions of the time \(t\) over the range \(0 \leq t \leq 5\) sec. Determine the maximum values of \(\theta\) and \(\dot{\theta}\) and the corresponding values of \(t .\) Also determine the first time at which \(\theta=90^{\circ}\)

A particle \(P\) is launched from point \(A\) with the initial conditions shown. If the particle is subjected to aerodynamic drag, compute the range \(R\) of the particle and compare this with the case in which aerodynamic drag is neglected. Plot the trajectories of the particle for both cases. Use the values \(v_{0}=\) \(65 \mathrm{m} / \mathrm{s}, \theta=35^{\circ},\) and \(k=4.0 \times 10^{-3} \mathrm{m}^{-1}\). (Note: The acceleration due to aerodynamic drag has the form \(\mathbf{a}_{D}=-k v^{2} \mathbf{t},\) where \(k\) is a positive constant, \(v\) is the particle speed, and \(t\) is the unit vector associated with the instantaneous velocity \(\mathbf{v}\) of the particle. The unit vector \(t\) has the form \(t=\frac{v_{x} \mathbf{i}+v_{y} \mathbf{j}}{\sqrt{v_{x}^{2}+v_{y}^{2}}},\) where \(v_{x}\) and \(v_{y}\) are the instantaneous \(x\) -and \(y\) -components of particle velocity, respectively.)
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