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Chapter 2: Problem 81

A particle's acceleration along an \(x\) axis is \(a=5.0 t\), with \(t\) in seconds and \(a\) in meters per second squared. At \(t=2.0 \mathrm{~s}\), its velocity is \(+17 \mathrm{~m} / \mathrm{s}\). What is its velocity at \(t=4.0 \mathrm{~s}\) ?

Short Answer

Expert verified
The velocity at \(t = 4.0\, \text{s}\) is \(47 \text{ m/s}\).

Step by step solution

01

Understand the Problem

A particle is moving along an x-axis with varying acceleration given by the function \(a = 5.0t\). We need to determine the velocity at \(t = 4.0\, \text{s}\), given that at \(t = 2.0\, \text{s}\), the velocity is \(+17 \text{m/s}\).
02

Integrate to Find Velocity Function

To find the velocity, we integrate the acceleration function. The acceleration is given by \(a = \frac{dv}{dt} = 5.0t\). Therefore, integrate with respect to \(t\):\[ v(t) = \int 5.0t \, dt = 2.5t^2 + C \]Here, \(C\) is a constant of integration representing the initial condition.
03

Determine Constant of Integration

We know that at \(t = 2.0\, \text{s}\), the velocity \(v = 17 \text{m/s}\). Using this information, substitute \(t = 2.0\) and \(v = 17\) into the velocity function:\[ 17 = 2.5(2.0)^2 + C \]Solving this, we find:\[ 17 = 2.5 \times 4 + C \]\[ 17 = 10 + C \]\[ C = 7 \]
04

Write Complete Velocity Function

Substitute \(C = 7\) into the velocity function to get the complete expression:\[ v(t) = 2.5t^2 + 7 \]
05

Calculate Velocity at t = 4.0 s

Substitute \(t = 4.0\) into the velocity function to find the velocity at this time:\[ v(4.0) = 2.5(4.0)^2 + 7 \]Calculate:\[ v(4.0) = 2.5 \times 16 + 7 \]\[ v(4.0) = 40 + 7 \]\[ v(4.0) = 47 \text{ m/s} \]

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Key Concepts

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

Integration
When dealing with kinematics, integration is a key mathematical tool used to find quantities like velocity and position from acceleration. In the given problem, the acceleration of the particle is expressed as a function of time, \(a = 5.0t\). Integration helps us find the velocity function from this acceleration. By integrating the acceleration function with respect to time, we can determine how the velocity changes over time.
  • The process begins by setting up the integral of the acceleration: \(v(t) = \int 5.0t \, dt \).
  • After integrating, we get: \(v(t) = 2.5t^2 + C\), where \(C\) is the constant of integration.
Initial conditions will help us determine the value of this constant \(C\). This integration process exemplifies how we move from acceleration to velocity in motion problems.
Acceleration
Acceleration is the rate at which an object's velocity changes with time. In kinematics, understanding acceleration is crucial for predicting how an object will move. In this exercise, the acceleration of the particle is given by the function \(a = 5.0t\), signifying a time-dependent acceleration.
  • This means that as time progresses, the acceleration increases linearly.
  • The factor of \(5.0\) indicates how quickly the acceleration itself changes as time advances.
A positive acceleration value, such as in this problem, suggests that the particle's velocity is increasing with time. This is integral to finding how fast and in what direction the particle is moving at a certain time.
Velocity
Velocity is a vector quantity that describes the speed of an object and its direction of motion. Understanding velocity is essential for predicting the future position of an object in kinematics. In this problem, once the function for velocity is obtained through integration, it is represented by:
  • \(v(t) = 2.5t^2 + 7\), after determining the constant of integration.
This equation tells us how the velocity of the particle changes with time. At \(t = 4.0\) seconds, we substitute that value into the velocity equation:\[v(4.0) = 2.5(4.0)^2 + 7 = 47 \, \text{m/s}\]The velocity is positive, indicating that the particle is moving forward along the x-axis, and it confirms that the speed is increasing as expected since the acceleration is also positive.
Initial Conditions
Initial conditions are specific values that are used to determine unknown constants in integrated functions. They are crucial in solving real-world motion problems. In this exercise, the problem provides an initial condition: the velocity at \(t = 2.0\) seconds is \(17 \, \text{m/s}\).
  • This information allows us to solve for the constant \(C\) in the integrated velocity function.
  • By substituting \(t = 2.0\) and \(v = 17\) into the equation \(v(t) = 2.5t^2 + C\), we resolve \(C = 7\).
With \(C\) determined, the complete velocity function becomes \(v(t) = 2.5t^2 + 7\). Without such initial conditions, the solution would remain incomplete, as the function could not be uniquely defined.

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Most popular questions from this chapter

A train started from rest and moved with constant acceleration. At one time it was traveling \(30 \mathrm{~m} / \mathrm{s}\), and \(160 \mathrm{~m}\) farther on it was traveling \(50 \mathrm{~m} / \mathrm{s}\). Calculate (a) the acceleration, (b) the time required to travel the \(160 \mathrm{~m}\) mentioned, \((\mathrm{c})\) the time required to attain the speed of \(30 \mathrm{~m} / \mathrm{s}\), and \((\mathrm{d})\) the distance moved from rest to the time the train had a speed of \(30 \mathrm{~m} / \mathrm{s}\). (e) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the train, from rest.

The head of a rattlesnake can accelerate at \(50 \mathrm{~m} / \mathrm{s}^{2}\) in striking a victim. If a car could do as well, how long would it take to reach a speed of \(100 \mathrm{~km} / \mathrm{h}\) from rest?

The speed of a bullet is measured to be \(640 \mathrm{~m} / \mathrm{s}\) as the bullet emerges from a barrel of length \(1.20 \mathrm{~m}\). Assuming constant acceleration, find the time that the bullet spends in the barrel after it is fired.

A pilot flies horizontally at \(1300 \mathrm{~km} / \mathrm{h}\), at height \(h=35 \mathrm{~m}\) above initially level ground. However, at time \(t=0\), the pilot begins to fly over ground sloping upward at angle \(\theta=4.3^{\circ}\) (Fig. \(2-41\) ). If the pilot does not change the airplane's heading, at what time \(t\) does the plane strike the ground?

Two subway stops are separated by \(1100 \mathrm{~m}\). If a subway train accelerates at \(+1.2 \mathrm{~m} / \mathrm{s}^{2}\) from rest through the first half of the distance and decelerates at \(-1.2 \mathrm{~m} / \mathrm{s}^{2}\) through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph \(x, v\), and \(a\) versus \(t\) for the trip.
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