The position of a passenger train that is traveling at an initial speed of 14 feet per second and continues to accelerate can be modeled by the function y = 14t2. A second train that is 1,400 feet ahead of the first train is traveling at a constant speed of 96 feet per second and can be modeled by the function y = 96t + 1400. Solve the system of equations. Which solution represents a viable time that the trains are side by side?

To find the time at which the two trains are side by side, we need to solve the system of equations given by the positions of the trains as functions of time \( t \):

1. \( y = 14t^2 \)
2. \( y = 96t + 1400 \)

Set the two equations equal to each other to find the time \( t \) when the trains are at the same position:
\[ 14t^2 = 96t + 1400 \]

Rearrange the equation to set it to zero:
\[ 14t^2 - 96t - 1400 = 0 \]

Next, solve this quadratic equation. We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 14 \), \( b = -96 \), and \( c = -1400 \).

First, calculate the discriminant:
\[ \Delta = b^2 - 4ac \]
\[ \Delta = (-96)^2 - 4(14)(-1400) \]
\[ \Delta = 9216 + 78400 \]
\[ \Delta = 87616 \]

Now, apply the quadratic formula:
\[ t = \frac{-(-96) \pm \sqrt{87616}}{2 \cdot 14} \]
\[ t = \frac{96 \pm 296}{28} \]

This gives us two potential solutions for \( t \):
\[ t = \frac{96 + 296}{28} = \frac{392}{28} = 14 \]
\[ t = \frac{96 - 296}{28} = \frac{-200}{28} = -\frac{100}{14} \approx -7.14 \]

Since time cannot be negative, we discard \( t \approx -7.14 \) and accept \( t = 14 \).

Therefore, the viable time at which the trains are side by side is \( \boxed{14 \text{ seconds}} \).

Do you want more details or have any questions?

Here are 5 additional questions for you to consider:
1. How do you interpret the coefficients in the quadratic formula?
2. What is the significance of the discriminant in a quadratic equation?
3. How do you solve a quadratic equation by completing the square?
4. Can quadratic equations have complex solutions? If so, when?
5. What are the real-world applications of solving systems of quadratic equations?

**Tip:** When solving real-world problems, always check that your solutions are feasible within the context of the problem. For example, negative time values might not be meaningful in certain scenarios.

Learn how to solve a system of equations involving two moving trains. Understand the quadratic formula and its application to real-world problems in physics and mathematics. Suitable for advanced high school students (Grades 10-12).

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