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A particle's acceleration is described by the function \(a_{x}=(10-t) \mathrm{m} / \mathrm{s}^{2},\) where \(t\) is in \(\mathrm{s}\). Its initial conditions are \(x_{0}=0 \mathrm{m}\) and \(v_{\mathrm{ax}}=0 \mathrm{m} / \mathrm{s}\) at \(t=0 \mathrm{s}\) a. At what time is the velocity again zero? b. What is the particle's position at that time?

Short Answer

Expert verified
a. The velocity of the particle is again zero at 20 seconds. b. The particle's position at that time is 1000 meters.

Step by step solution

01

Find Velocity Function

Given the acceleration function \( a_{x}=(10-t)\,m/s^{2}\),we can find the velocity function \( v_{x} \) by integrating the acceleration function. Let's take the integral \( \int a(t)\, dt \) from 0 to \( t \), using the fact that \( v_{0}=0 \) at \( t=0 \). This yields \( v_{x}=10t-\frac{1}{2}t^{2} \).
02

Determine Time when Velocity is Zero

Now that we have the velocity function, we can find the time when the velocity is again zero by setting \( v_{x} \) equal to zero and solving for \( t \). Doing so gives us the equation \( 10t-\frac{1}{2}t^{2}=0 \). Solving this for \( t \) yields \( t=20 \) seconds.
03

Find Position Function

Now to find the position. As velocity is the derivative of position, we can integrate the velocity function to get the position. Taking the integral \( \int v(t)\, dt \) from 0 to \( t \), using \( x_{0}=0 \) at \( t=0 \) gives us the position function \( x_{x}=5t^{2}-\frac{1}{6}t^{3} \).
04

Determine Position when Velocity is Zero

Finally, to answer question b, we input the value of \( t = 20 \) (as obtained in step 2) into the position function. Doing so we get \( x_{20}=5(20)^{2}-\frac{1}{6}(20)^{3}= 1000 \) meters.

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Key Concepts

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

Velocity-Time Integration
When studying motion, an important concept is how velocity changes over time when a particle is under constant acceleration. Velocity-time integration is a method used to find an object's velocity at a specific time by integrating its acceleration function with respect to time. Imagine pushing a sled down a snowy hill; the sled will accelerate due to gravity. To calculate how fast it's moving after a certain time, we could use velocity-time integration.

For instance, given an acceleration function, like the textbook problem suggests, you start with an initial velocity and integrate acceleration to find the new velocity. In our example, the initial conditions were zero velocity, and by integrating the acceleration function from time zero to any given time (t), you're essentially summing up all the tiny changes in velocity that occur every moment, which results in the overall change in velocity from the starting point to time (t).
Position-Time Integration
Position-time integration takes the process a step further by determining the position of an object after it has been in motion for a while. The position of an object is affected by both its initial position and its velocity over time. By integrating the velocity function with respect to time, we get the position function.

Consider a toy car rolling across the floor; to determine how far it has traveled after accelerating for a few seconds, we'd integrate its velocity over that time period. Using the velocity calculated from the acceleration function, we integrate from the initial position (which was zero in the textbook problem) to find how far the particle moves. It's like calculating the area under the velocity-time curve, which represents the distance the particle has traveled during the given time.
Acceleration Function
The acceleration function describes how a particle's acceleration changes over time. It is a key component in understanding motion because it provides the rate at which the velocity changes. In a practical sense, an acceleration function could be used to model a car's increasing speed as the driver pushes on the gas pedal.

In our exercise, the acceleration function is given as a linear function of time, meaning the acceleration decreases linearly as time progresses. It is crucial to correctly understand and apply the acceleration function as it is the foundation for finding the velocity and position functions through integration. It encapsulates the essence of the particle's dynamic behavior over the time interval.
Kinematics Equations
Kinematics equations are the formulas used to describe the motion of an object without considering the forces that cause the motion. These equations relate the variables of motion - position, velocity, acceleration, and time - in ways that allow us to predict an object's future state or reconstruct its past movements.

In the realm of kinematics, there are typically four main equations, each of which applies under different conditions. In the context of our textbook problem, we don't have constant acceleration, which means we cannot use the standard kinematics equations directly. Instead, integration becomes a critical tool to derive the velocity and position as functions of time when dealing with a variable acceleration. The equations derived from these functions are specific cases of the broader kinematics equations adapted for our problem's unique conditions.

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Most popular questions from this chapter

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40 mph and half the distance at 60 mph. On her return trip, she drives half the time at 40 mph and half the time at 60 mph. a. What is Julie's average speed on the way to Grandmother's house? b. What is her average speed on the return trip?

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The minimum stopping distance for a car traveling at a speed of \(30 \mathrm{m} / \mathrm{s}\) is \(60 \mathrm{m},\) including the distance traveled during the driver's reaction time of \(0.50 \mathrm{s}\) a. What is the minimum stopping distance for the same car traveling at a speed of \(40 \mathrm{m} / \mathrm{s} ?\) b. Draw a position-versus-time graph for the motion of the car in part a. Assume the car is at \(x_{0}=0\) m when the driver first sees the emergency situation ahead that calls for a rapid halt.

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equation $$v_{x}^{2}=\frac{2 P}{m} t$$ where \(P=3.6 \times 10^{4}\) watts is the car's power output, \(m=\) \(1200 \mathrm{kg}\) is its mass, and \(v_{x}\) is in \(\mathrm{m} / \mathrm{s} .\) That is, the square of the car's velocity increases linearly with time. a. What is the car's speed at \(t=10 \mathrm{s}\) and at \(t=20 \mathrm{s} ?\) b. Find a symbolic expression, in terms of \(P, m,\) and \(t,\) for the car's acceleration at time \(\iota\) c. Evaluate the acceleration at \(t=1 \mathrm{s}\) and \(t=10 \mathrm{s}\) d. This simple model fails for \(t\) less than about \(0.5 \mathrm{s}\). Explain how you can recognize the failure. e. Find a symbolic expression for the distance the car has traveled at time \(t\) f. One-quarter mile is \(402 \mathrm{m}\). What is the Spider's best time in a quarter-mile race? (The model's failure in the first 0.5 s has very little effect on your answer because the car travels almost no distance during that time.)

A hotel elevator ascends \(200 \mathrm{m}\) with a maximum speed of \(5.0 \mathrm{m} / \mathrm{s} .\) Its acceleration and deceleration both have a magnitude of \(1.0 \mathrm{m} / \mathrm{s}^{2}\) a. How far does the elevator move while accelerating to full speed from rest? b. How long does it take to make the complete trip from bottom to top?

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