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Chapter 2: Problem 70

Two particles move along an \(x\) axis. The position of particle 1 is given by \(x=6.00 t^{2}\) \(+3.00 t+2.00\) (in meters and seconds); the acceleration of particle 2 is given by \(a=-8.00 t\) (in meters per second squared and seconds) and, at \(t=0\), its velocity is \(20 \mathrm{~m} / \mathrm{s}\). When the velocities of the particles match, what is their velocity?

Short Answer

Expert verified
The velocity of both particles when they match is approximately 15.6 m/s.

Step by step solution

01

Find Velocity Function of Particle 1

To find the velocity of particle 1, we need to differentiate its position with respect to time:\[ v_1 = \frac{d}{dt}(6.00t^2 + 3.00t + 2.00) = 12.00t + 3.00. \]So, the velocity of particle 1 is \(v_1 = 12.00t + 3.00\) m/s.
02

Find Velocity Function of Particle 2

To determine the velocity of particle 2, we first integrate the acceleration function with respect to time. Given \(a = -8.00t\), integrate to get the velocity:\[ v_2 = \int -8.00 t \, dt = -4.00t^2 + C. \]With initial velocity \(v_2(0) = 20.00\, \text{m/s}\), we find the constant \(C\):\[ v_2(0) = -4.00(0)^2 + C = 20.00 \Rightarrow C = 20.00. \]Thus, the velocity of particle 2 is \(v_2 = -4.00t^2 + 20.00\) m/s.
03

Set Velocity Functions Equal

To find the time \(t\) when both particles have the same velocity, set their velocity equations equal to each other:\[ 12.00t + 3.00 = -4.00t^2 + 20.00. \]Rearrange to form a quadratic equation:\[ 4.00t^2 + 12.00t + 3.00 - 20.00 = 0 \Rightarrow 4.00t^2 + 12.00t - 17.00 = 0. \]
04

Solve Quadratic Equation for Time

Solve the quadratic equation \(4.00t^2 + 12.00t - 17.00 = 0\) using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.00\), \(b = 12.00\), and \(c = -17.00\):\[ t = \frac{-12.00 \pm \sqrt{12.00^2 - 4 \times 4.00 \times (-17.00)}}{2 \times 4.00}. \]Calculate the discriminant:\[ \sqrt{12.00^2 + 272.00} = \sqrt{416.00}. \]Thus:\[ t = \frac{-12.00 \pm 20.39}{8.00}. \]This yields two potential times:\[ t_1 = 1.05 \quad \text{and} \quad t_2 = -4.55. \]Since time cannot be negative, \(t = 1.05\) seconds.
05

Find Velocity at Meeting Time

Using the time \(t = 1.05\) s, substitute back into either velocity function (they should be equal):\[ v = 12.00 \times 1.05 + 3.00 = 15.60 \text{ m/s} \approx 15.6 \text{ m/s}. \]

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Key Concepts

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

Velocity
Velocity refers to how fast an object is moving and in what direction. It is a vector quantity, meaning it has both magnitude and direction. In the context of our problem, the velocity of particle 1 depends on the change in position over time. By differentiating the position function of particle 1, which is given as \( x = 6.00t^2 + 3.00t + 2.00 \), we find its velocity function as \( v_1 = 12.00t + 3.00 \) meters per second. This means for every second, the velocity changes linearly with time at a rate of 12 meters per second, with an initial velocity of 3 meters per second when \( t = 0 \).
  • Remember, velocity tells us not just how fast, but also in which direction something is moving.
  • Units of velocity typically include meters per second \( m/s \) or kilometers per hour \( km/h \).
Acceleration
Acceleration is the rate of change of velocity with time. It is another vector quantity which means it shows how quickly velocity itself is changing. In our problem, the acceleration for particle 2 is given by \( a = -8.00t \). This negative acceleration value tells us that the particle is slowing down as time goes on, meaning it's decelerating.
  • If acceleration is constant, such as in uniform acceleration, velocity changes at a steady rate.
  • Negative acceleration, or deceleration, implies the object is slowing down.
  • The units for acceleration are typically meters per second squared (\( m/s^2 \)).
Quadratic Equation
Quadratic equations are polynomial equations of degree 2, generally of the form \( ax^2 + bx + c = 0 \). They are pivotal in kinematics when describing motion with constant acceleration. To find when both particles have the same velocity, we converted our velocity match into a quadratic equation: \( 4.00t^2 + 12.00t - 17.00 = 0 \).
  • Solutions to quadratic equations can be found using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • The discriminant \( b^2 - 4ac \) determines the nature of the solutions (real and distinct, real and repeated, or complex).
  • Quadratic equations are critical in solving various physical problems, especially those involving paths of objects under uniform acceleration.
Differentiation
Differentiation is a calculus technique used to find the rate at which one quantity changes with respect to another. In our problem, it's used to find the velocity of particle 1 from its position function. By differentiating \( x = 6.00t^2 + 3.00t + 2.00 \), we obtained the velocity function as \( v_1 = 12.00t + 3.00 \).
  • Differentiation provides a way to determine velocity from position, or acceleration from velocity.
  • The derivative of a constant is zero, as constants do not change.
  • In physics, differentiation helps compute instantaneous rates like velocity and acceleration at any given point.
Integration
Integration is a fundamental calculus operation that is essentially the reverse process of differentiation. It helps us find a function when its rate of change is known. For particle 2, we integrated the acceleration \( a = -8.00t \) to find its velocity function \( v_2 = -4.00t^2 + C \).
  • Integration can be thought of as finding the area under a curve, which combines small elements to form a whole.
  • It is used to derive position from velocity or velocity from acceleration when initial conditions are known.
  • Initial conditions, such as \( v_2(0) = 20.00 \) m/s, are necessary for integrating to determine the constant term.

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Most popular questions from this chapter

A key falls from a bridge that is \(45 \mathrm{~m}\) above the water. It falls directly into a model boat, moving with constant velocity, that is \(12 \mathrm{~m}\) from the point of impact when the key is released. What is the speed of the boat?

If the position of a particle is given by \(x=20 t-5 t^{3}\), where \(x\) is in meters and \(t\) is in seconds, when, if ever, is the particle's velocity zero? (b) When is its acceleration \(a\) zero? (c) For what time range (positive or negative) is \(a\) negative? (d) Positive? (e) Graph \(x(t), v(t)\), and \(a(t)\).

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m} ;\) both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\) Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

An electric vehicle starts from rest and accelerates at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) in a straight line until it reaches a speed of \(20 \mathrm{~m} / \mathrm{s}\). The vehicle then slows at a constant rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop? -26 A muon (an elementary particle) enters a region with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) and then is slowed at the rate of \(1.25 \times\) \(10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) (a) How far does the muon take to stop? (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the muon.
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