Problem 20 - Algebra Form and Function

Question

A car's distance (in miles) from home after \(t\) hours is given by \(s(t)=11 t^{2}+t+40\). (a) How far from home is the car at \(t=0 ?\) (b) Use function notation to express the car's position after 1 hour and then find its position. (c) Use function notation to express the statement "For what value of \(t\) is the car 142 miles from home?" (d) Write an equation whose solution is the time when the car is 142 miles from home. (e) Use trial and error for a few values of \(t\) to determine when the car is 142 miles from home.

Step-by-Step Solution

Answer

Answer: The car is approximately 142 miles away from home between \(t=3\) and \(t=4\).

Step 1: (a) Car's position at t = 0

To find the car's position at t = 0, we need to evaluate the function \(s(t)\) at t=0. In other words, replace the value of \(t\) in the equation with 0: \(s(0)=11(0)^2+(0)+40\).

Step 2: (a) Calculation for t = 0

Since \(t=0\), let's calculate \(s(0)\): \(s(0) = 11(0)^2 + (0) + 40 = 0 + 0 + 40 = 40\) miles. So, the car is 40 miles away from home at \(t=0\).

Step 3: (b) Car's position after 1 hour

To find the car's position after 1 hour, we need to evaluate the function \(s(t)\) at t=1. In other words, replace the value of \(t\) in the equation with 1: \(s(1)=11(1)^2+(1)+40\).

Step 4: (b) Calculation for t = 1

Since \(t=1\), let's calculate \(s(1)\): \(s(1) = 11(1)^2 + (1) + 40 = 11 + 1 + 40 = 52\) miles. So, the car is 52 miles away from home after 1 hour.

Step 5: (c) Function notation for 142 miles away

To express the statement "For what value of \(t\) is the car 142 miles from home?" using function notation, set the function equal to 142 miles: \(s(t) = 142\).

Step 6: (d) Equation for when the car is 142 miles away

We need to write an equation for the time when the car is 142 miles away from home. We can use the given formula \(s(t)=11t^{2}+t+40\) and set it equal to 142: \(11t^{2} + t + 40 = 142\).

Step 7: (e) Trial and error for t

Now let's use trial and error for different values of \(t\) to see when the car is 142 miles away from home: For \(t = 2\): \(s(2) = 11(2)^2 + (2) + 40 = 96\) For \(t = 3\): \(s(3) = 11(3)^2 + (3) + 40 = 130\) For \(t = 4\): \(s(4) = 11(4)^2 + (4) + 40 = 180\) As we can see, when \(t=3\), the car is 130 miles away from home, and when \(t=4\), the car is 180 miles away from home. Since 142 miles is between 130 and 180, we can estimate that the car is 142 miles away from home between \(t=3\) and \(t=4\). To get a more precise answer, further calculations can be done using more refined values of \(t\) between 3 and 4, or other methods such as solving the quadratic equation can be employed.

Key Concepts

Understanding Quadratic Equations

A quadratic equation is an invaluable tool in mathematics, commonly expressed in the form \( ax^2 + bx + c = 0 \). It's a second-degree polynomial equation, meaning it contains the variable raised to the power of two. In many real-world scenarios, such as calculating trajectories or optimizing resource use, quadratic equations help model complex activities.
In our exercise, the car's distance from home is modeled using a quadratic equation: \( s(t) = 11t^2 + t + 40 \). This signifies that the car's distance varies as a function of the square of time and linear time. The coefficients (11, 1, and 40) determine how the distance changes over time. Quadratic equations can be solved in various ways, such as factoring, completing the square, or using the quadratic formula.
Understanding how to work with these equations enables us to predict and extrapolate many real-life occurrences, like in our scenario, predicting the car's distance over a period.

Applying the Distance Formula

The distance formula used in the context of this exercise allows us to calculate how far the car is from home at any point in time \(t\). It is modeled by the function \( s(t) = 11t^2 + t + 40 \). This is an example of a function where distance \(s(t)\) is expressed as a function of time (\(t\)).
The distance formula is crucial because:

• It provides an exact measure of how far the car travels from its starting point at time \(t=0\).
• By evaluating \(s(t)\) for specific values of \(t\), you can determine the distance at those given times.

In our example, when \(t=0\), the car is initially 40 miles away. With the chosen use of function notation, this expression helps simplify and illustrate changes over time and distance, making it easier to analyze movement trends.

Mastering Trial and Error Method

The trial and error approach is a hands-on method often employed when evaluating a function for which an exact solution is difficult to find. It involves guessing various values for \(t\), evaluating the function at each, and refining guesses based on the results.
In the exercise, we used trial and error to estimate when the car is exactly 142 miles from home by testing different \(t\) values. Initially:

• For \(t = 2\), the car is 96 miles away.
• For \(t = 3\), it is 130 miles away.
• For \(t = 4\), the distance is 180 miles.

The goal is to get closer to 142 miles. With these results, it becomes clear that \(t\) is between 3 and 4. The method shines when more refined intervals are chosen, further pinpointing the accurate \(t\) value. While basic, trial and error remains a potent technique for honing in on a solution, especially when other methods are challenging to apply.