Chapter 1: Problem 43
Solving for a Variable Solve the equation for the indicated variable. $$h=\frac{1}{2} g t^{2}+v_{0} t ; \quad \text { for } t$$
Short Answer
Expert verified
The solutions for \( t \) are \( t = \frac{-v_{0} \pm \sqrt{v_{0}^{2} + 2gh}}{g} \); choose the positive root.
Step by step solution
01
Analyze the given equation
The equation given is \( h = \frac{1}{2} g t^{2} + v_{0} t \). We are asked to solve for \( t \). This means we need to rearrange the equation to express \( t \) explicitly in terms of the other variables \( h, g, \) and \( v_{0} \).
02
Reformulate as a standard quadratic equation
Rearrange the equation to fit the standard quadratic form \( at^2 + bt + c = 0 \). The equation already resembles this form: \( \frac{1}{2}gt^2 + v_{0}t - h = 0 \). Here, \( a = \frac{1}{2}g \), \( b = v_{0} \), and \( c = -h \).
03
Apply the quadratic formula
The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) allows us to solve for \( t \). Substitute \( a = \frac{1}{2}g \), \( b = v_{0} \), and \( c = -h \) into the formula. This gives:\[ t = \frac{-v_{0} \pm \sqrt{v_{0}^{2} - 4 \times \frac{1}{2}g \times (-h)}}{2 \times \frac{1}{2}g} \].
04
Simplify the expression
First, compute the discriminant: \[ b^2 - 4ac = v_{0}^{2} + 2gh \].Next, substitute back to get:\[ t = \frac{-v_{0} \pm \sqrt{v_{0}^{2} + 2gh}}{g} \].
05
Interpret the results
The solution \( t = \frac{-v_{0} + \sqrt{v_{0}^{2} + 2gh}}{g} \) and \( t = \frac{-v_{0} - \sqrt{v_{0}^{2} + 2gh}}{g} \) provides two potential values for \( t \). Depending on the context (e.g., time can't be negative), we select the appropriate root \( t = \frac{-v_{0} + \sqrt{v_{0}^{2} + 2gh}}{g} \) assuming it is positive.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving for a Variable
Solving for a variable means finding the value of one variable in an equation relative to the other variables. It’s like trying to solve a mystery by finding out what one unknown value equals. In equations such as the quadratic ones, solving for a variable involves getting it by itself on one side of the equation.
Here, we were tasked with solving for the variable \( t \) in the equation \( h = \frac{1}{2} g t^{2} + v_{0} t \). This involves using algebraic techniques like rearranging and substituting values to get \( t \) alone.
Key strategies often include:
Here, we were tasked with solving for the variable \( t \) in the equation \( h = \frac{1}{2} g t^{2} + v_{0} t \). This involves using algebraic techniques like rearranging and substituting values to get \( t \) alone.
Key strategies often include:
- Identifying the term with the variable you're solving for.
- Rearranging the equation, possibly involving other variables if necessary.
- Using mathematical formulas tailored to the structure of the equation.
- Simplifying calculations to obtain the desired expression for the variable.
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the second degree. A typical quadratic equation can be written as \( at^2 + bt + c = 0\). The quadratic formula \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] allows us to find the values of \( t \) that satisfy this equation.
To use the formula effectively, identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation. In our specific exercise, \( a = \frac{1}{2}g \), \( b = v_{0} \), and \( c = -h \). Substitute these values into the quadratic formula to solve for \( t \). This technique is incredibly helpful for determining the roots of any quadratic equation.
To use the formula effectively, identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation. In our specific exercise, \( a = \frac{1}{2}g \), \( b = v_{0} \), and \( c = -h \). Substitute these values into the quadratic formula to solve for \( t \). This technique is incredibly helpful for determining the roots of any quadratic equation.
- Remember to check the discriminant \( b^2 - 4ac \) before applying the formula, as it informs the number and type of solutions.
- The solutions from the formula can help visualize possible values for real-world contexts, like finding time in motion problems.
Discriminant
The discriminant is a component of the quadratic formula and is key to understanding the nature of the solutions of a quadratic equation. The discriminant, represented by \( b^2 - 4ac \), determines:
- The number of real roots of the equation.
- The nature of these roots, whether they are real or complex.
- If \( b^2 - 4ac > 0 \), there are two distinct real roots.
- If \( b^2 - 4ac = 0 \), there’s exactly one real root (a repeated root).
- If \( b^2 - 4ac < 0 \), the roots are complex, not real.
Rearranging Equations
Rearranging equations is a fundamental mathematical skill that involves altering the structure of an equation to make it easier to solve for a specific variable. It's like rearranging pieces of a puzzle to see the picture you need.
In our exercise, the original equation \( h = \frac{1}{2} g t^{2} + v_{0} t \) was rearranged into the standard quadratic form \( \frac{1}{2}gt^2 + v_{0}t - h = 0 \). This step was crucial because it set up the equation for the application of the quadratic formula.
Here are some general tips when rearranging:
In our exercise, the original equation \( h = \frac{1}{2} g t^{2} + v_{0} t \) was rearranged into the standard quadratic form \( \frac{1}{2}gt^2 + v_{0}t - h = 0 \). This step was crucial because it set up the equation for the application of the quadratic formula.
Here are some general tips when rearranging:
- Perform the same operation on both sides to maintain equality.
- Keep track of positive and negative signs, especially when moving terms across the equals sign.
- Factor and simplify terms where possible to reduce complexity.
- Double-check each step to ensure that the rearrangement is mathematically sound.