Problem 40 - Calculus Early Transcendentals

Question

Let \(s_{A}=15 t^{2}+10 t+20\) and \(s_{B}=5 t^{2}+40 t, t \geq 0,\) be the position functions of cars \(A\) and \(B\) that are moving along parallel straight lanes of a highway. (a) How far is car \(A\) ahead of car \(B\) when \(t=0\) ? (b) At what instants of time are the cars next to each other? (c) At what instant of time do they have the same velocity? Which car is ahead at this instant?

Step-by-Step Solution

Answer

(a) 20 units; (b) At \( t=1 \) and \( t=2 \); (c) At \( t=1.5 \), Car A is ahead.

Step 1: Calculate the position of Car A at t=0

To find how far Car A is ahead when \( t = 0 \), we need to evaluate Car A's position function: \( s_A = 15t^2 + 10t + 20 \). Substituting \( t = 0 \) gives \( s_A = 15(0)^2 + 10(0) + 20 = 20 \).

Step 2: Calculate the position of Car B at t=0

Similarly, we calculate Car B's position by evaluating its position function: \( s_B = 5t^2 + 40t \). Substituting \( t = 0 \) gives \( s_B = 5(0)^2 + 40(0) = 0 \).

Step 3: Determine distance when t=0

At \( t = 0 \), Car A is at position 20 and Car B is at position 0. Therefore, Car A is \( 20 \) units ahead of Car B.

Step 4: Set position functions equal to find when cars are next to each other

To find when the cars are next to each other, set \( s_A = s_B \): \( 15t^2 + 10t + 20 = 5t^2 + 40t \). Simplify to \( 10t^2 - 30t + 20 = 0 \).

Step 5: Solve the quadratic equation

Solve the equation \( 10t^2 - 30t + 20 = 0 \) using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 10, b = -30, c = 20 \). Simplify to find \( t \).

Step 6: Analyze the solutions for time

The quadratic formula gives \( t = \frac{30 \pm \sqrt{900 - 800}}{20} = \frac{30 \pm 10}{20} \) so \( t = 2 \) or \( t = 1 \). The cars are next to each other at these times.

Step 7: Differentiate to find velocities

The velocities are the derivatives of the position functions. For Car A: \( v_A = \frac{d}{dt}(15t^2 + 10t + 20) = 30t + 10 \). For Car B: \( v_B = \frac{d}{dt}(5t^2 + 40t) = 10t + 40 \).

Step 8: Set velocity functions equal for the same speed

Set the velocities equal to find when the speeds are the same: \( 30t + 10 = 10t + 40 \). Solve for \( t \) to get \( 20t = 30 \), thus \( t = 1.5 \).

Step 9: Evaluate positions at t=1.5 to see which car is ahead

Substitute \( t = 1.5 \) into both position functions. For Car A: \( 15(1.5)^2 + 10(1.5) + 20 = 68.75 \). For Car B: \( 5(1.5)^2 + 40(1.5) = 67.5 \). Car A is ahead at \( t = 1.5 \).

Key Concepts

Position Functions

Position functions are essential in calculus for understanding how the location of an object, such as a car on a highway, changes over time. These functions, typically denoted as \( s(t) \), express the position based on time \( t \).

• For example, the position function for Car A is given by \( s_A = 15t^2 + 10t + 20 \), while for Car B, it's \( s_B = 5t^2 + 40t \).
• The coefficients of these equations dictate how quickly the position changes with respect to time.
• Here, \( 15t^2 \) and \( 5t^2 \) represent the acceleration effects over time squared, whereas the linear terms \( 10t \) and \( 40t \), represent consistent speed influences.

Hence, by evaluating these position functions at various time points \( t \), we can determine where each car is located. For instance, at \( t = 0 \), the starting positions of the vehicles are checked to find how far apart they are initially.

Quadratic Equation

A quadratic equation is fundamental in calculus and algebra, typically expressed in the form \( ax^2 + bx + c = 0 \). It is applied in many real-world scenarios, including analyzing motion in position functions.

• In our exercise, we set the position functions of Car A and Car B equal to each other to determine when the cars meet. Hence, the equation \( 15t^2 + 10t + 20 = 5t^2 + 40t \) simplifies to the quadratic form \( 10t^2 - 30t + 20 = 0 \).
• This can be solved using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), providing the potential times when the cars are side by side.
• Solving yields \( t = 2 \) or \( t = 1 \), indicating that at these times, the positions of Car A and Car B are the same.

Solving quadratic equations is crucial for finding key intersections or events in position-based problems.

Velocity Derivatives

Deriving position functions to find their derivatives gives us the velocity functions, which are the rate of change of position concerning time. Understanding these derivatives forms the backbone of motion analysis in calculus.

• The derivative of a position function is its velocity. For Car A, differentiating \( s_A = 15t^2 + 10t + 20 \) gives the velocity function \( v_A = 30t + 10 \).
• Similarly, for Car B, \( s_B = 5t^2 + 40t \) leads to \( v_B = 10t + 40 \).
• By setting the derived velocity functions equal, \( 30t + 10 = 10t + 40 \), we solve for \( t = 1.5 \). This tells us the time at which both cars have the same speed.
• This process highlights how calculus helps compare motion dynamics across different objects in motion, providing insights that are not immediately obvious from position functions alone.

Velocity derivatives allow us to determine important dynamics, like when cars moving differently will match speeds, reflecting deeper insights into their movements.