Carissa is crossing a very, very, very wide street, and did not properly check both ways before doing so. (Don't be like Carissa!) She initially begins walking at $2$ feet per second. Suddenly, she hears a car approaching, and begins running, eventually making it safely to the other side, half a minute after she began crossing. Given that Carissa always runs $n$ times as fast as she walks and that she spent $n$ times as much time running as she did walking, and given that the street is $260$ feet wide, find Carissa's running speed, in feet per second.

[i]Proposed by Andrew Wu[/i]
1. Let \( a \) be the number of seconds Carissa spent walking. According to the problem, she spent \( n \) times as much time running as she did walking. Therefore, the time she spent running is \( an \) seconds.
2. The total time she spent crossing the street is given as 30 seconds. Thus, we have the equation:
   \[
   a + an = 30
   \]
   Factoring out \( a \) from the left-hand side, we get:
   \[
   a(1 + n) = 30
   \]
   Solving for \( a \), we find:
   \[
   a = \frac{30}{1 + n}
   \]
3. Carissa walks at a speed of 2 feet per second. Therefore, the distance she walked is:
   \[
   2a \text{ feet}
   \]
   Since she runs \( n \) times as fast as she walks, her running speed is \( 2n \) feet per second. The distance she ran is:
   \[
   2n \cdot an \text{ feet} = 2an^2 \text{ feet}
   \]
4. The total distance she covered is the width of the street, which is 260 feet. Therefore, we have the equation:
   \[
   2a + 2an^2 = 260
   \]
   Dividing the entire equation by 2, we get:
   \[
   a + an^2 = 130
   \]
5. Substituting \( a = \frac{30}{1 + n} \) into the equation \( a + an^2 = 130 \), we get:
   \[
   \frac{30}{1 + n} + \frac{30n^2}{1 + n} = 130
   \]
   Combining the terms on the left-hand side, we have:
   \[
   \frac{30(1 + n^2)}{1 + n} = 130
   \]
   Multiplying both sides by \( 1 + n \), we get:
   \[
   30(1 + n^2) = 130(1 + n)
   \]
   Expanding both sides, we have:
   \[
   30 + 30n^2 = 130 + 130n
   \]
   Rearranging the terms, we get:
   \[
   30n^2 - 130n - 100 = 0
   \]
6. This is a quadratic equation in the form \( an^2 + bn + c = 0 \). We can solve it using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 30 \), \( b = -130 \), and \( c = -100 \):
   \[
   n = \frac{130 \pm \sqrt{(-130)^2 - 4 \cdot 30 \cdot (-100)}}{2 \cdot 30}
   \]
   Simplifying inside the square root:
   \[
   n = \frac{130 \pm \sqrt{16900 + 12000}}{60}
   \]
   \[
   n = \frac{130 \pm \sqrt{28900}}{60}
   \]
   \[
   n = \frac{130 \pm 170}{60}
   \]
   This gives us two potential solutions:
   \[
   n = \frac{300}{60} = 5 \quad \text{or} \quad n = \frac{-40}{60} = -\frac{2}{3}
   \]
   Since \( n \) must be a positive number, we discard \( n = -\frac{2}{3} \) and take \( n = 5 \).

7. Carissa's running speed is \( 2n \) feet per second. Therefore, her running speed is:
   \[
   2 \cdot 5 = 10 \text{ feet per second}
   \]

The final answer is \( \boxed{10} \) feet per second.