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Chapter 2: Problem 93

Two cars, \(A\) and \(B\) , travel in a straight line. The distance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

Short Answer

Expert verified
Both cars start at the same point. They are at the same position at \(t = 0\) and another time found using the quadratic formula. Distance is neither increasing nor decreasing when \(\alpha + 2\beta t = 2\gamma t - 3\delta t^2\). They have the same acceleration when \(t = \frac{2\gamma - 2\beta}{6\delta}\).

Step by step solution

01

Find Initial Positions

To determine which car is initially ahead, evaluate the positions of both cars, \(x_A(t)\) and \(x_B(t)\), at \(t = 0\). We find \(x_A(0) = \alpha \cdot 0 + \beta \cdot 0^2 = 0\) and \(x_B(0) = \gamma \cdot 0^2 - \delta \cdot 0^3 = 0\). Thus, both cars start at the same position.
02

Find Equal Position Time(s)

Solve for time when the distances of both cars are equal: \(x_A(t) = x_B(t)\). Set \(\alpha t + \beta t^2 = \gamma t^2 - \delta t^3\) and rearrange to yield the equation \(- \delta t^3 + (\beta - \gamma) t^2 + \alpha t = 0\). Factoring out \(t\), we get \(t(- \delta t^2 + (\beta - \gamma)t + \alpha) = 0\). Consequently, \(t = 0\) or solve \(- \delta t^2 + (\beta - \gamma)t + \alpha = 0\). Using the quadratic formula, \(t = \frac{-(\beta - \gamma) \pm \sqrt{(\beta - \gamma)^2 - 4(-\delta)(\alpha)}}{2(-\delta)}\), we calculate \(t\).
03

Calculate Derivative for Increasing/Decreasing Distance

To determine when the distance between the cars is neither increasing nor decreasing, find \(\frac{d}{dt}(x_A(t) - x_B(t))\). First, differentiate to get the velocities: \(v_A(t) = \frac{d}{dt}(x_A(t)) = \alpha + 2\beta t\) and \(v_B(t) = \frac{d}{dt}(x_B(t)) = 2\gamma t - 3\delta t^2\). Set \(v_A(t) - v_B(t) = 0\) to find when velocities are equal, which simplifies to \(\alpha + 2\beta t - 2\gamma t + 3\delta t^2 = 0\). Solve this equation for \(t\).
04

Find Time(s) of Equal Acceleration

The accelerations of the cars are obtained by differentiating their velocities. For \(A\), \(a_A = 2\beta\), and for \(B\), \(a_B = 2\gamma - 6\delta t\). Set \(a_A = a_B\) to find \(t\): \(2\beta = 2\gamma - 6\delta t\), thus \(t = \frac{2\gamma - 2\beta}{6\delta}\).

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Key Concepts

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

Quadratic Equations
In kinematics, quadratic equations are incredibly useful for describing motion, especially when dealing with acceleration. Let's explore this further using the expressions for the distances traveled by cars A and B:

  • For car A: \(x_A(t) = \alpha t + \beta t^2\)
  • For car B: \(x_B(t) = \gamma t^2 - \delta t^3\)
Both equations include terms that involve time \(t\) squared (and for car B, even time cubed), making them polynomial equations of degree two or higher. This structure helps represent the changing speeds due to acceleration.

To find when the cars are at the same position, we set these equations equal and solve for \(t\). Simplifying gives us a quadratic equation:

\[- \delta t^3 + (\beta - \gamma)t^2 + \alpha t = 0\]

Solving it efficiently often requires the quadratic formula, commonly used when equations cannot be factored easily. This process of solving quadratic equations is essential for comparing positions and analyzing motion in physics.
Derivatives in Physics
Derivatives play a pivotal role in physics, especially when analyzing motion. They allow us to understand how quantities change over time, providing insights into speed (velocity) and how speed itself changes (acceleration).

For our problem, we need derivatives to find the velocities of cars A and B:
  • Velocity of car A, \(v_A(t) = \frac{d}{dt}(x_A(t)) = \alpha + 2\beta t\)
  • Velocity of car B, \(v_B(t) = \frac{d}{dt}(x_B(t)) = 2\gamma t - 3\delta t^2\)
These are first derivatives of the position functions. Setting the velocities equal allows us to find when the distance between cars A and B is neither increasing nor decreasing. Derivatives provide critical insights in physics homework as they help us explore the dynamic aspects of moving objects.
Acceleration Analysis
Understanding acceleration is key to fully grasping kinematics, as it describes how the speed of an object changes over time. In the given exercise, the accelerations for cars A and B are derived by taking the second derivative of the position functions, or the first derivative of their respective velocity functions.
  • Acceleration of car A: \(a_A = \frac{d}{dt}(v_A(t)) = 2\beta\)
  • Acceleration of car B: \(a_B = \frac{d}{dt}(v_B(t)) = 2\gamma - 6\delta t\)
These formulas help us determine at what time \(t\) the accelerations of A and B are equal. Set them equal: \(2\beta = 2\gamma - 6\delta t\) to solve for \(t\). This gives insights into how each car's speed is changing relative to the other, a crucial element in kinematic analysis.

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Most popular questions from this chapter

A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the uverage speed and (b) the average velocity of the marble.

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of \(384,000 \mathrm{km}\) . Suppose the spaceship starts from rest and accelerates at 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) for the first 15.0 min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) , just coming to rest as it reaches the moon. (a) What is the maximum speed attained? (b) What fraction of the total distance is traveled at constant speed? (c) What total time is required for the trip?

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