Problem 39 - Calculus Early Transcendentals

Question

Suppose that the position functions of two particles, \(P_{1}\) and \(P_{2},\) in motion along the same line are $$ s_{1}=\frac{1}{2} t^{2}-t+3 \quad \text { and } \quad s_{2}=-\frac{1}{4} t^{2}+t+1 $$ $$ \begin{array}{l}{\text { respectively, for } t \geq 0} \\ {\text { (a) Prove that } P_{1} \text { and } P_{2} \text { do not collide. }} \\ {\text { (b) How close do } P_{1} \text { and } P_{2} \text { get to each other? }} \\\ {\text { (c) During what intervals of time are they moving in oppo- }} \\\ {\text { site directions? }}\end{array} $$

Step-by-Step Solution

Answer

(a) They do not collide. (b) Closest at \( t = \frac{8}{3} \), distance \( \approx 0.889 \). (c) Moving opposite for \( t \in (1, 2) \).

Step 1: Determine When P1 and P2 Collide

To find if the particles collide, we set their position functions equal: \( s_1 = s_2 \). This gives the equation \( \frac{1}{2}t^2 - t + 3 = -\frac{1}{4}t^2 + t + 1 \). Simplifying, we have \( \frac{3}{4}t^2 - 2t + 2 = 0 \). Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = \frac{3}{4}, b = -2, c = 2 \), we find the discriminant \( b^2 - 4ac = 4 - 6 = -2 \). Since the discriminant is negative, the quadratic has no real roots, so the particles do not collide.

Step 2: Defining How Close P1 and P2 Get

The distance between particles at any time \( t \) is given by \( d(t) = |s_1 - s_2| = \left| \frac{3}{4}t^2 - 2t + 2 \right| \). This is a quadratic function opening upwards, reaching its minimum at the vertex \( t = -\frac{b}{2a} = \frac{2}{\frac{3}{4}} = \frac{8}{3} \). However, since \( t \) must be non-negative, evaluate \( d(t) \) at 0 to find it's within boundaries in [0, 8/3]. Substitute, \( d(0) = 2 \), the minimum \( d(\frac{8}{3}) = \left( \frac{8}{9} \right) \approx 0.889 \). Thus, the closest they get is approximately 0.889 units.

Step 3: Analyzing Motion Directions

The particles move in opposite directions when the derivatives of their position functions, \( v_1(t) = s_1'(t) \) and \( v_2(t) = s_2'(t) \), have opposite signs. Compute the velocities: \( v_1(t) = \frac{d}{dt} \left(\frac{1}{2}t^2 - t + 3 \right) = t - 1 \) and \( v_2(t) = \frac{d}{dt} \left(-\frac{1}{4}t^2 + t + 1 \right) = -\frac{1}{2}t + 1 \). These are opposite for \( (t-1)(-\frac{1}{2}t+1) < 0 \). Solves to find intervals, solving each as \( t - 1 > 0 \) and \( -\frac{1}{2}t + 1 > 0 \): \( t > 1 \) and \( t < 2 \). The intervals where \( t \in (1,2) \) they move in opposite directions.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In this exercise, the position functions of the particles involve quadratic equations. To determine whether the particles \( P_1 \) and \( P_2 \) collide, we equate their position functions and simplify to the quadratic equation \( \frac{3}{4}t^2 - 2t + 2 = 0 \). Solving quadratic equations often requires using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The key component here is the discriminant \( b^2 - 4ac \), which indicates the number and type of roots.

• If the discriminant is positive, the equation has two distinct real roots.
• If it's zero, there is exactly one real root, indicating the vertex of the parabola lies on the x-axis.
• If the discriminant is negative, as in this exercise, the equation has no real roots. This means \( P_1 \) and \( P_2 \) will not collide at any point in time. Hence, quadratic equations help determine the behavior and interaction between particles in motion.

Particle Motion

Particle motion can be described using position functions that provide a mathematical representation of the particles' locations over time. For particles moving along a straight line, these functions are typically dependent on time \( t \). In this exercise, the position functions are quadratic, indicated by terms involving \( t^2 \). Such functions reflect the path taken by the particles over the time interval \( t \ge 0 \).
By calculating the difference in these position functions, we can analyze how close the particles \( P_1 \) and \( P_2 \) get to each other. The distance function \( d(t) = |s_1 - s_2| \) helps us identify their relative positions. Determining the vertex of this quadratic gives us the minimum distance between the two particles. In this case, evaluating \( d(t) \) reveals that the minimum distance they reach is approximately 0.889 units. This kind of analysis is crucial for understanding particle interaction with respect to time.

Velocity Analysis

Velocity analysis in physics involves determining a particle's speed and direction of motion. It's derived from the derivative of the position function, often termed the velocity function \( v(t) \). Here, for \( P_1 \), the velocity \( v_1(t) = t - 1 \) and for \( P_2 \), the velocity \( v_2(t) = -\frac{1}{2}t + 1 \). By calculating these derivatives, we gain insights into how fast and in what direction the particles move.

• When the velocity is positive, the particle moves forward.
• When negative, the particle moves backward.

To find when \( P_1 \) and \( P_2 \) move in opposite directions, we analyze when one velocity function is positive while the other is negative. Solving \((t - 1)(-\frac{1}{2}t + 1) < 0\) helps identify the interval \( t \in (1,2) \) where the velocities differ in sign, indicating the particles are moving opposite each other. Understanding these intervals is essential for predicting the trajectories and possible interactions of moving particles.