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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 velocities of both particles match at 15.00 m/s.

Step by step solution

01

Identify Position Function for Particle 1

The position of particle 1 along the x-axis is given by the quadratic function \(x = 6.00t^2 + 3.00t + 2.00\). To find the velocity, we need to differentiate this function with respect to time \(t\).
02

Calculate Velocity of Particle 1

The velocity \(v_1\) of particle 1 is the first derivative of the position function with respect to time:\[v_1(t) = \frac{d}{dt}(6.00t^2 + 3.00t + 2.00) = 12.00t + 3.00\]
03

Develop Velocity Function for Particle 2

The acceleration function for particle 2 is given as \(a = -8.00t\). We integrate this function to find the velocity function. Using the initial condition \(v_2(0) = 20\, m/s\):\[v_2(t) = \int -8.00t\, dt = -4.00t^2 + C\]. Applying the initial condition at \(t=0\), we find \(C = 20\), so \[v_2(t) = -4.00t^2 + 20\].
04

Solve for Time when Velocities are Equal

Set the expressions for the velocities of particles 1 and 2 equal to each other to find the time when they have the same velocity:\[12.00t + 3.00 = -4.00t^2 + 20\]. Rearranging terms, we get the quadratic equation:\[4.00t^2 + 12.00t - 17 = 0\].
05

Solve the Quadratic Equation

Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.00\), \(b = 12.00\), and \(c = -17\):\[t = \frac{-12.00 \pm \sqrt{12.00^2 - 4 \cdot 4.00 \cdot (-17)}}{2 \cdot 4.00}\]\[t = \frac{-12.00 \pm \sqrt{144 + 272}}{8}\]\[t = \frac{-12.00 \pm 20.00}{8}\]Solving gives two potential solutions, \(t = 1.00\, s\) or \(t = -4.25\, s\). Since time cannot be negative, \(t = 1.00\, s\) is valid.
06

Calculate Velocity at Time when Velocities Match

Substitute \(t = 1.00\, s\) into either velocity equation, we'll use particle 1's:\[v_1(1.00) = 12.00(1.00) + 3.00 = 15.00\, m/s\].

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Key Concepts

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

Understanding Velocity
Velocity in physics describes how quickly an object is moving and in which direction. It's a vector quantity, which means it includes both magnitude and direction.
For particle 1, the velocity is derived from its position function by differentiation. The position function given is quadratic, indicating that the velocity changes over time. This change is observed by taking the derivative:
  • The function is: \(x=6.00 t^{2}+3.00 t+2.00\).
  • Differentiating with respect to time \(t\), gives us the velocity: \(v_1(t) = 12.00t + 3.00\).
This equation indicates that as time increases, the velocity of particle 1 changes linearly in time.For particle 2, velocity isn't given directly. Instead, we have acceleration, and integrating it gives us velocity. Initially, particle 2 starts with a velocity of 20 m/s.
Insight into Acceleration
Acceleration is the rate at which velocity changes with time. It tells us how quickly an object is speeding up or slowing down.
In the case of particle 2, its acceleration is given by a linear function of time:
  • Acceleration function: \(a = -8.00t\).
The negative sign indicates that particle 2 is decelerating, slowing down as time increases.
To determine velocity from acceleration, integration is necessary. This concept contrasts with differentiation done for particle 1, where velocity is computed from position. For particle 2's motion, integrating acceleration gives:
  • Velocity function: \(v_2(t) = -4.00t^2 + C\), where \(C\) is a constant determined by an initial condition.
  • The initial velocity \(v_2(0) = 20 m/s\) provides the value for this constant: \(C = 20\).
Exploring Position Function
The position function represents an object's location on a defined path over time. It's essential for tracking movement and understanding how far and in what direction an object travels.
For particle 1, the position is described by a quadratic function:
  • Position function: \(x=6.00 t^{2}+3.00 t+2.00\).
The terms in this function each influence movement differently:
  • \(6.00t^2\) indicates acceleration impact from the start.
  • \(3.00t\) shows initial linear motion contribution.
  • \(2.00\) shows the starting position or shift along the x-axis.
Position function points to where the object will be at any time \(t\), which is instrumental for subsequent velocity calculations.
The Role of Differentiation
Differentiation in calculus is the process of finding the rate at which a function is changing at any point. In kinematics, differentiation is primarily used to transition from a position to a velocity function.Here's how it works for particle 1:
  • Given position function: \(x = 6.00t^2 + 3.00t + 2.00\).
  • To find velocity, we differentiate the position function with respect to time:\(v_1(t) = \frac{d}{dt}(6.00t^2 + 3.00t + 2.00) = 12.00t + 3.00\).
Differentiation provides the means to progress in understanding an object's motion. It allows us to calculate how movement changes at any given moment. This tool is crucial in solving physics problems and gaining insights into dynamic systems.

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Most popular questions from this chapter

At a certain time a particle had a speed of \(18 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction, and \(2.4 \mathrm{~s}\) later its speed was \(30 \mathrm{~m} / \mathrm{s}\) in the opposite direction. What is the average acceleration of the particle during this \(2.4 \mathrm{~s}\) interval?

Water drips from the nozzle of a shower onto the floor \(200 \mathrm{~cm}\) below. The drops fall at regular (equal) intervals of time. the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nocrle are the (a) second and (b) third drops?

If a baseball pitcher throws a fastball at a horizontal speed of \(160 \mathrm{~km} / \mathrm{h},\) how long does the ball take to reach home plate \(18.4 \mathrm{~m}\) away?

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building \(1.60 \mathrm{~s}\) after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground \(6.00 \mathrm{~s}\) after it is launched. In \(\mathrm{SI}\) units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

Suppose a rocket ship in deep space moves with con- stant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed onc-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?\) (b) How far will it travel in so doing?
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