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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?

Short Answer

Expert verified
(a) 20 units; (b) at \(t = 1\) and \(t = 2\); (c) at \(t = 1.5\), car B is ahead.

Step by step solution

01

Calculating Initial Distance Between Cars

At time \( t = 0 \), calculate the positions using the given functions \( s_A(t) = 15t^2 + 10t + 20 \) and \( s_B(t) = 5t^2 + 40t \). Substitute \( t = 0 \) to find \( s_A(0) = 15(0)^2 + 10(0) + 20 = 20 \) and \( s_B(0) = 5(0)^2 + 40(0) = 0 \). Thus, car A is 20 units ahead of car B.
02

Setting Up Equations for Equal Positions

To find when the cars are next to each other, solve the equation \( s_A(t) = s_B(t) \). This means solving \( 15t^2 + 10t + 20 = 5t^2 + 40t \). Simplify this to \( 10t^2 - 30t + 20 = 0 \).
03

Solving the Quadratic Equation

Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( 10t^2 - 30t + 20 = 0 \), where \( a = 10 \), \( b = -30 \), and \( c = 20 \). Calculate \( t = \frac{30 \pm \sqrt{(-30)^2 - 4*10*20}}{2*10} = \frac{30 \pm \sqrt{100}}{20} \). Therefore, \( t = 2 \text{ or } 1 \).
04

Finding Velocities of Cars A and B

Compute the velocities \( v_A(t) \) and \( v_B(t) \) by differentiating the position functions. \( v_A(t) = \frac{d}{dt}(15t^2 + 10t + 20) = 30t + 10 \) and \( v_B(t) = \frac{d}{dt}(5t^2 + 40t) = 10t + 40 \).
05

Equating Velocities to Find Instant of Same Velocity

Set \( 30t + 10 = 10t + 40 \) to find \( t \). Simplify to \( 20t = 30 \), giving \( t = 1.5 \). At \( t = 1.5 \), calculate \( s_A(1.5) = 15(1.5)^2 + 10(1.5) + 20 \) and \( s_B(1.5) = 5(1.5)^2 + 40(1.5) \) to determine that car B is ahead.

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Key Concepts

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

Quadratic Equations
Quadratic equations are a fundamental concept in algebra and calculus. These equations are typically expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In solving these problems, understanding the nature of quadratic equations helps predict the trajectory of moving objects.
Quadratic equations can have zero, one, or two real solutions depending on the value of the discriminant, \( b^2 - 4ac \).
  • If the discriminant is positive, the equation has two real and distinct solutions.
  • If zero, there is one real solution, often called a repeated or double root.
  • If negative, the solutions are complex numbers.
In the exercise, we set the position functions of cars equal to each other to find when they are next to each other. This involves solving the quadratic equation \( 10t^2 - 30t + 20 = 0 \), which provides the times when the cars have equal positions on the highway.
Velocity
Velocity measures the rate of change of an object's position over time, providing insight into how fast it's moving and in which direction. It's a key concept in physics and calculus. The velocity function is the first derivative of the position function.
The exercise involves finding the velocities of car A and car B from their position functions. Differentiating the position function of car A, \( s_A(t) = 15t^2 + 10t + 20 \), gives the velocity \( v_A(t) = 30t + 10 \). Similarly, differentiating the position function for car B, \( s_B(t) = 5t^2 + 40t \), gives \( v_B(t) = 10t + 40 \).
To find when both cars have the same velocity, we equate the two velocity functions and solve for time \( t \). This results in finding at \( t = 1.5 \), both cars are moving at the same rate.
Position Functions
Position functions describe the location of a moving object at any given time. These functions vary with time and help determine both the exact spot of the object and its path over time.
In the original exercise, car A's position function is \( s_A(t) = 15t^2 + 10t + 20 \) while car B's is \( s_B(t) = 5t^2 + 40t \). Position functions in this form reveal both the initial position and the changing behavior over time due to the time-dependent components.
  • The square term \( t^2 \) indicates acceleration because it involves time squared.
  • The linear term \( t \) represents a constant velocity contribution.
  • The constant represents the starting position at \( t = 0 \).
By comparing the position functions, you can determine the relative location of the two cars over time. For example, setting these two equal helps find the time instances where the cars are level with each other on the highway.

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