Problem 51
Question
An object travels on a horizontal line. The distance it travels is represented by \(d\) and is measured in meters. The equation relating time of travel, \(t,\) and distance of travel, \(d\), is \(d=t^{2}-4 t+20\) Determine the distance traveled by the object if it has been in motion for 6 seconds.
Step-by-Step Solution
Verified Answer
Answer: The object has traveled 32 meters after 6 seconds.
1Step 1: Identify the given information
We know the time for which the object has been in motion, \(t = 6\) seconds, and the equation of motion, \(d = t^{2} - 4t + 20\).
2Step 2: Substitute the given time into the equation
Plug in the value of \(t = 6\) into the equation of motion to find the distance, \(d\): \(d = (6)^{2} - 4(6) + 20\).
3Step 3: Solve for the distance traveled
By solving the equation, we can find the distance traveled at \(t = 6\) seconds:
\(d = (6)^{2} - 4(6) + 20\)
\(d = 36 - 24 + 20\)
\(d = 12 + 20\)
\(d = 32\)
So, the distance traveled by the object after 6 seconds is 32 meters.
Key Concepts
Quadratic EquationsDistance-time RelationshipAlgebraic Problem-solvingSubstitution Method
Quadratic Equations
Quadratic equations are a fundamental aspect of algebra that often describe the motion of objects in physics. They have the general form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable that we are trying to solve. These equations can have two real solutions, one real solution, or no real solutions depending on the value of the discriminant \(b^2 - 4ac\).
For example, in the context of our exercise, the equation representing the distance travelled by an object, \(d = t^{2} - 4t + 20\), is indeed a quadratic equation in terms of time \(t\). The terms \(t^{2}\), \(-4t\), and \(20\) represent the coefficients analogous to \(a\), \(b\), and \(c\), respectively. Identifying the structure of a quadratic equation can aid in predicting the behavior of an object's travel over time and is crucial to solving many real-world problems.
For example, in the context of our exercise, the equation representing the distance travelled by an object, \(d = t^{2} - 4t + 20\), is indeed a quadratic equation in terms of time \(t\). The terms \(t^{2}\), \(-4t\), and \(20\) represent the coefficients analogous to \(a\), \(b\), and \(c\), respectively. Identifying the structure of a quadratic equation can aid in predicting the behavior of an object's travel over time and is crucial to solving many real-world problems.
Distance-time Relationship
The distance-time relationship in physics is an essential concept for understanding how objects move. It's typically expressed as a function or an equation showing how the distance an object travels changes over time. In simple scenarios, you might see a linear relationship where distance increases steadily as time goes on. However, motion can also be more complex, requiring quadratic equations to describe it accurately.
In our exercise example, the relationship between time and distance is quadratic, which indicates that the object’s speed is changing as time passes. The equation \(d = t^{2} - 4t + 20\) reveals that the distance doesn't just depend on time, but on time squared and time itself, hinting at an acceleration component. Understanding this relationship allows us to calculate the position of an object at any given time.
In our exercise example, the relationship between time and distance is quadratic, which indicates that the object’s speed is changing as time passes. The equation \(d = t^{2} - 4t + 20\) reveals that the distance doesn't just depend on time, but on time squared and time itself, hinting at an acceleration component. Understanding this relationship allows us to calculate the position of an object at any given time.
Algebraic Problem-solving
Algebraic problem-solving is a process involving steps to simplify and solve expressions and equations. It is foundational to success in many fields of study, including mathematics, physics, engineering, and economics. Good problem-solving skills allow one to identify relevant information, establish the relationship between different quantities, and apply mathematical operations to find solutions.
In our exercise, we begin by identifying the motion equation and the given time. By following an ordered approach - translating the problem into an equation, identifying known and unknowns, and performing algebraic manipulations - we can solve for the distance travelled after a specified time. This process is crucial as it develops a systematic approach to handle more complex scenarios and problems encountered in various academic disciplines.
In our exercise, we begin by identifying the motion equation and the given time. By following an ordered approach - translating the problem into an equation, identifying known and unknowns, and performing algebraic manipulations - we can solve for the distance travelled after a specified time. This process is crucial as it develops a systematic approach to handle more complex scenarios and problems encountered in various academic disciplines.
Substitution Method
The substitution method is a technique used in algebra to solve for the unknown variables in an equation. This method involves replacing a variable with its corresponding value, which simplifies the equation and makes it easier to solve. It is particularly useful when dealing with equations that describe real-life situations, such as the motion of an object.
In the exercise provided, the substitution method is applied by taking the known value of time \(t = 6\) seconds and substituting it into the motion equation. This simplifies the equation to one that can be solved using basic arithmetic, leading us to the distance travelled by the object. The substitution method provides a clear and direct path to the solution, minimizing the complexity of dealing with variables and abstract terms.
In the exercise provided, the substitution method is applied by taking the known value of time \(t = 6\) seconds and substituting it into the motion equation. This simplifies the equation to one that can be solved using basic arithmetic, leading us to the distance travelled by the object. The substitution method provides a clear and direct path to the solution, minimizing the complexity of dealing with variables and abstract terms.
Other exercises in this chapter
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