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Chapter 5: Problem 53

Set up appropriate systems of two linear equations and solve the systems algebraically. All data are accurate to at least two significant digits. In a test of a heat-seeking rocket, a first rocket is launched at \(2000 \mathrm{ft} / \mathrm{s},\) and the heat-seeking rocket is launched along the same flight path 12 s later at a speed of \(3200 \mathrm{ft} / \mathrm{s}\). Find the times \(t_{1}\) and \(t_{2}\) of flight of the rockets until the heat- seeking rocket destroys the first rocket.

Short Answer

Expert verified
The flight times are \( t_1 = 32 \) seconds and \( t_2 = 20 \) seconds.

Step by step solution

01

Identify the Problem

We are given two rockets with different speeds and start times. The goal is to determine how long each rocket will be in flight until the heat-seeking rocket reaches the first rocket. We assume both rockets travel the same distance by the time they meet.
02

Define Variables

Let \( t_1 \) be the time in seconds that the first rocket is in flight until it is destroyed, and let \( t_2 \) be the time in seconds that the heat-seeking rocket is in flight until it meets the first rocket. We know \( t_2 = t_1 - 12 \) since the heat-seeking rocket is launched 12 seconds later.
03

Set Up Two Equations

The distance travelled by each rocket can be expressed as the product of speed and time.For the first rocket: \( 2000t_1 \).For the heat-seeking rocket: \( 3200t_2 \).Since both travel the same distance by interception, we have the equation: \( 2000t_1 = 3200t_2 \).
04

Solve for t2 in Terms of t1

From \( 2000t_1 = 3200t_2 \), express \( t_2 \) in terms of \( t_1 \) by dividing both sides by 3200: \( t_2 = \frac{2000}{3200}t_1 \). Simplify this to \( t_2 = \frac{5}{8}t_1 \).
05

Substitute t2 Expression into Timing Equation

Using the relationship \( t_2 = t_1 - 12 \), substitute the expression for \( t_2 \) from the previous step:\[ \frac{5}{8}t_1 = t_1 - 12 \].
06

Solve For t1

Rearrange this equation to solve for \( t_1 \):\( \frac{5}{8}t_1 - t_1 = -12 \) which simplifies to \( -\frac{3}{8}t_1 = -12 \).Multiply both sides by \(-\frac{8}{3}\) to isolate \( t_1 \):\( t_1 = 32 \).
07

Calculate t2

Use the relationship \( t_2 = t_1 - 12 \) to find \( t_2 \):\( t_2 = 32 - 12 = 20 \).
08

Verify the Solution

Check that both rockets travel the same distance:First rocket: \( 2000 \times 32 = 64000 \) feet.Heat-seeking rocket: \( 3200 \times 20 = 64000 \) feet.Both distances are equal, confirming the solution is correct.

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Key Concepts

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

Linear Equations
Linear equations are one of the foundational concepts in algebra. A linear equation is an equation where the highest power of the variable is one. These equations are represented in the form of \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants. In solving physics and kinematics problems, like the one involving rockets, linear equations help us describe the relationship between different factors such as time, speed, and distance.
  • In this problem, we use the linear equation to relate speed and time to distance.
  • The general equation for distance is \( ext{distance} = ext{speed} imes ext{time} \).
Understanding and forming linear equations turn physical scenarios into solvable mathematical problems. They serve as the stepping stones to finding solutions in various scientific applications.
Algebraic Solutions
The term algebraic solutions refers to solving equations or sets of equations using algebraic manipulations. In the case of simultaneous linear equations, this involves utilizing substitution or elimination methods to find the values of unknown variables. Let's take our rocket problem as an example.
  • We set up the equation based on the problem statement: \( 2000t_1 = 3200t_2 \).
  • This algebraic approach requires expressing one variable in terms of another, allowing us to solve for each variable step-by-step.
In solving these equations, clarity and logical progression are key. By using algebraic manipulations like simplifying equations and substituting variables, we arrive methodically at the required answers. It is through these processes that algebra bridges the gap between conceptual theory and practical application.
Physics Problems
Physics problems often involve understanding the laws of nature and applying mathematical formulas to find solutions. The problem with the rockets demonstrates how physics intertwines with mathematics, particularly algebra.
  • It begins with interpreting real-world problems into mathematical terms, making use of given data like speeds and time intervals.
  • By framing the problem with linear equations, we translate kinetic scenarios into solvable equations.
Grasping physics problems through such exercises enhances your ability to apply theoretical physics concepts through mathematical frameworks, marrying observation and calculation in practical problem-solving.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces causing the motion. In solving problems like the rocket exercise, we use kinematic equations that express relationships among time, distance, and velocity.
  • The kinematic equation, \( ext{distance} = ext{speed} imes ext{time} \) is fundamental for these calculations.
  • By understanding when and how to utilize such equations, scenarios become easier to visualize and solve mathematically.
Kinematics allows us to focus on the motion's outcomes 鈥� distance travelled in our rocket example 鈥� helping dissect complex motion dynamics into simpler, comprehensible terms.

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Most popular questions from this chapter

Solve the given problems by determinants. In Exerciser-46,\( set up appropriate systems of equations. All numbers are accurate to at least two significant digits. A company budgets \)\$ 750,000$ in salaries, hardware, and computer time for the design of a new product. The salaries are as much as the others combined, and the hardware budget is twice the computer budget. How much is budgeted for each?

In order to make the coefficients easier to work with, first multiply each term of the equation or divide each term of the equation by a number selected by inspection. Then proceed with the solution of the system by an appropriate algebraic method. $$\begin{array}{l} 250 R+225 Z=400 \\ 375 R-675 Z=325 \end{array}$$

Solve the given problems by determinants. In a laboratory experiment to measure the acceleration of an object, the distances traveled by the object were recorded for three different time intervals. These data led to the following equations: $$\begin{array}{l} s_{0}+2 v_{0}+2 a=20 \\ s_{0}+4 v_{0}+8 a=54 \\ s_{0}+6 v_{0}+18 a=104 \end{array}$$ Here, \(s_{0}\) is the initial displacement (in \(\mathrm{ft}\) ), \(v_{0}\) is the initial velocity (in \(\mathrm{ft} / \mathrm{s})\), and \(a\) is the acceleration (in \(\mathrm{ft} / \mathrm{s}^{2}\) ). Find \(s_{0}, v_{0},\) and \(a\).

Evaluate the given third-order determinants. $$\left|\begin{array}{rrr} 8 & 9 & -6 \\ -3 & 7 & 2 \\ 4 & -2 & 5 \end{array}\right|$$

Use the determinant at the right. Answer the questions about the determinant for the changes given in each exercise. \(\left|\begin{array}{lll}4 & 2 & 1 \\\ 3 & 6 & 5 \\ 7 & 9 & 8\end{array}\right|=19\) How does the value change if the elements of the first row are added to the corresponding elements of the second row (the first row remains unchanged)?
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