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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 m/s and \(\beta =\) 1.20 m/s\(^2\). The distance of \(B\) from the starting point is \(x_B(t) = \gamma{t}^2 - \delta{t}^3\), with \(\gamma =\) 2.80 m/s\(^2\) and \(\delta =\) 0.20 m/s\(^3\). (a) Which car is ahead just after the two cars 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
(a) Car A. (b) Solve: \( t = \frac{-1.60 \pm \, \text{sqrt}((1.60)^2 - 4(-0.20)(-2.60))}{2(-0.20)} \). (c) Solve: \( 0.60t^2 - 3.00t + 2.60 = 0 \). (d) \( t = \frac{3.20}{1.20} \).

Step by step solution

01

Initial Distance Comparison

To determine which car is ahead just after the start, we evaluate their distances at a small time, i.e., just after the start, let's calculate the value at \(t = 0^+\).\( x_A(0^+) = \alpha \cdot 0 + \beta \cdot 0^2 = 0 \text{ m}\)\( x_B(0^+) = \gamma \cdot 0^2 - \delta \cdot 0^3 = 0 \text{ m}\)At \(t = 0\), \(x_A = 0\) and \(x_B = 0\), hence they are at the same position. Just after this point, we need to evaluate the derivatives to see who starts moving away more rapidly.\(x_A'(t) = \alpha + 2\beta{t} = 2.60 + 2(1.20)t = 2.60 + 2.40t\)\(x_B'(t) = 2\gamma{t} - 3\delta{t}^2 = 2(2.80)t - 3(0.20)t^2 = 5.60t - 0.60t^2\)Initially, for a very small \(t\), \(x_A\) has a non-zero speed, while \(x_B\) does not move at \(t = 0\) since \(5.60t - 0.60t^2\approaches 5.60t\to 0 \, \text{ as } \, t\to 0\). So, Car \(A\) moves away first, hence Car \(A\) is initially ahead.
02

Find Intersection Time

To find when the cars meet again, solve for \(x_A(t) = x_B(t)\).\( \alpha{t} + \beta{t}^2 = \gamma{t}^2 - \delta{t}^3\)Substitute the values of \(\alpha, \beta, \gamma, \text{ and } \delta\):\( 2.60t + 1.20t^2 = 2.80t^2 - 0.20t^3\)Rearrange the equation:\( 0 = -0.20t^3 + 1.60t^2 - 2.60t\)Factor out the common factor of t:\( t(-0.20t^2 + 1.60t - 2.60) = 0\)So, the solutions are \(t = 0\) or solve the quadratic:\(-0.20t^2 + 1.60t - 2.60 = 0\)Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = -0.20, b = 1.60, c = -2.60\):\(t = \frac{-1.60 \pm \sqrt{(1.60)^2 - 4(-0.20)(-2.60)}}{2(-0.20)}\)Calculate to find the valid positive time values.
03

Distance Change Analysis

For the distance from A to B to neither increase nor decrease, the derivative of the distance needs to be zero.Distance \(D(t) = x_B(t) - x_A(t)\).\(D'(t) = x_B'(t) - x_A'(t)\)\(D'(t) = (5.60t - 0.60t^2) - (2.60 + 2.40t)\)Simplify and find when \(D'(t) = 0\):\(3.00t - 0.60t^2 = 2.60\)Rearrange and solve this quadratic:\(0.60t^2 - 3.00t + 2.60 = 0\)Use the quadratic formula again to find the time.
04

Acceleration Comparison

Find the accelerations and equate them to find when they are the same.Acceleration of Car A, \(a_A = x_A''(t) = 2\beta = 2.40\,\text{m/s}^2\).Acceleration of Car B, \(a_B = x_B''(t) = 2\gamma - 6\delta{t} = 5.60 - 1.20t\).Equate these:\(2.40 = 5.60 - 1.20t\)Solve for \(t\) to find when these accelerations are equal.\(1.20t = 5.60 - 2.40 = 3.20\)\(t = \frac{3.20}{1.20}\)Calculate to get the time value.

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Key Concepts

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

Equations of Motion
In physics, the equations of motion are fundamental relations that describe how objects move. These formulas allow us to predict the future positions, velocities, and accelerations of moving objects based on their present values and some properties of the force acting on them. Often applied in kinematics, these equations are derived from the basic assumptions of Newton's laws of motion.
At their core, these kinematic equations are:
  • Displacement from velocity and time: \( x = x_0 + v_0t + \frac{1}{2}at^2 \)
  • Velocity in terms of initial velocity, acceleration, and time: \( v = v_0 + at \)
  • Velocity in terms of initial velocity, acceleration, and displacement: \( v^2 = v_0^2 + 2a(x - x_0) \)
In the problem, Car A's position is described with a linear term \( \alpha t \) and a quadratic term \( \beta t^2 \), which correspond to initial velocity and constant acceleration components, respectively.
Meanwhile, Car B's equation has quadratic \( \gamma t^2 \) and cubic terms \( -\delta t^3 \), reflecting a changing acceleration state as time progresses.
Acceleration
Acceleration is defined as the rate of change of velocity with respect to time. It tells us how quickly an object is speeding up or slowing down. The concept of acceleration is crucial because it not only tells us about the speed change but also indicates the direction of motion change.

In physics, acceleration \( a \) is often expressed through the second derivative of the position with respect to time (or the first derivative of velocity), and it has the following formula:
  • \( a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \)
Car A has a constant acceleration given by the coefficient \( 2\beta \), which simplifies to 2.40 m/s². This means Car A's rate of speed increase is steady over time.

Car B’s acceleration is more complex, described by the expression \(2\gamma - 6\delta t\). This suggests Car B's acceleration changes over time because of the \(-6\delta t\) term, which reduces the acceleration as \( t \) increases, indicating a variable acceleration.
Quadratic Equations
Quadratic equations are key in describing the motion of objects because they appear naturally when an object's acceleration is constant. They take the standard form: \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( x \) is the unknown variable.

To solve these equations for \( x \), we often use the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our context, the quadratic equation helps calculate when two cars have the same position and when the distance between them does not change. This is achieved by equating their distances and solving: \(-0.20t^3 + 1.60t^2 - 2.60t = 0\).
Notably, finding the times using the quadratic formula allows us to identify intersections, crucial in understanding relative motion. Knowing how to handle these quadratic solutions is invaluable in predicting and analyzing moving objects in various scenarios.

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Most popular questions from this chapter

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