Question

The distance \(s_{1}\) traveled by one falling body in time \(t\) is given by \(3.74 t^{2}+5.83 t+4.22 \mathrm{ft}\) and the distance \(s_{2}\) traveled by another falling body in the same time is given by \(9.76 t^{2}+2.95 t+1.94 \mathrm{ft} .\) Write an expression for the distance \(s_{2}-s_{1}\) between the two bodies, and simplify.

Short answer

The simplified expression for the distance between two bodies as a function of time is \(s_2 - s_1 = 6.02t^2 - 2.88t - 2.28\) feet.

Step-by-step solution

Determine the Expression for Distance

Write down the given expression for the distances traveled by both bodies. Let's call these expressions as functions of time, i.e., for the first body, the distance function is represented as: \(s_1(t) = 3.74t^2 + 5.83t + 4.22\), and for the second body: \(s_2(t) = 9.76t^2 + 2.95t + 1.94\).

Set Up the Difference of Distances

Now, write an expression to find the difference in distance traveled by the two bodies as a function of time: \(s_2 - s_1 = (9.76t^2 + 2.95t + 1.94) - (3.74t^2 + 5.83t + 4.22)\).

Combine Like Terms

Simplify the difference by combining like terms: subtract the coefficients of \(t^2\), subtract the coefficients of \(t\), and subtract the constants from each other to get the simplified expression: \(s_2 - s_1 = (9.76 - 3.74)t^2 + (2.95 - 5.83)t + (1.94 - 4.22)\).

Simplify the Expression

Perform the subtractions to get the final, simplified expression: \(s_2 - s_1 = 6.02t^2 - 2.88t - 2.28\). This is the expression for the distance between the two bodies, simplified.

Key concepts

Quadratic Equations

At the core of many physics problems, including those involving the motion of falling bodies, are quadratic equations. They are mathematical expressions of the second degree, generally represented in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. These equations are pivotal in describing parabolic trajectories, which is exactly the type of path a body in free fall follows under the influence of gravity, ignoring air resistance.

Understanding the Coefficients

In the case of falling bodies, the coefficients have physical meanings: \( a \) is half the acceleration due to gravity, \( b \) can represent initial velocity, and \( c \) might be the initial height from which the body starts falling. In the exercise provided, \( s_1(t) \) and \( s_2(t) \) are quadratic equations representing the distances fallen by two bodies over time. By subtracting these equations, we tackle a real-world problem using a fundamental concept of algebra.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra that allows us to reduce equations to their simplest form, making them easier to interpret or solve. When we talk about simplifying expressions in the context of quadratic equations, like the ones we encounter in the exercise, we focus on combining like terms and performing operations to minimize the equation's complexity.

How to Simplify

The process of simplification often involves several steps: distributing any multiplication over addition or subtraction, combining like terms by adding or subtracting coefficients, and possibly factoring. In the textbook solution, simplifying the expression \( s_2 - s_1 \) involves subtracting the coefficients of the quadratic term \( t^2 \), the linear term \( t \) and the constant term to arrive at a much more straightforward expression that clearly depicts the changing distance between the two bodies over time.

Physics in Calculus

Physics often employs calculus to describe and analyze the motion and interaction of objects. In the case of falling bodies, calculus is utilized to interpret the objects’ changing positions, velocities, and accelerations over time. The functions that define displacement with respect to time are derivatives of velocity, and the acceleration functions are derivatives of these.

In our exercise, even though we didn't directly use differential calculus to solve the problem, understanding that the derivative of the distance functions \( s_1(t) \) and \( s_2(t) \) would give us velocity, and the derivative of the velocity function would yield acceleration, reinforces the connection between these physics concepts and calculus. The idea that quadratic equations can represent physical phenomena and their simplification is a stepping stone to more complex analyses involving derivatives and integrals in physics problems.