*(a) How far from the stop line do you pass the Porsche? *

At the instant you pass the Porsche, the x values of your car and the Porsche must be equal. Let's write the general equations for x(t):

General equation: x = x_{o} + v_{o} t + ½ a t^{2}

For your car : x_{you} = -15 + 11.11 t + 0

For the Porsche : x_{Porsche} = 0 + 0 +½ (3) t^{2} = 1.5 t^{2}

Now what?

Set x_{you} = x_{Porsche}, and solve for the time(s) when
the
two cars are at the same position. Then plug the time back into
the x
equation(s) to find the position.

x_{you} = x_{Porsche} leads to:
-15 + 11.11 t = 1.5 t^{2}

Bringing everything to one side gives:
1.5 t^{2} - 11.11 t + 15 = 0

Solve using the quadratic formula:

where a = 1.5, b = -11.11, and c = 15

This gives two values for t, t = 1.776 s and t = 5.631 s.

To answer question (a), plug t = 1.776 into either of your x expressions.
Both equations **must** give the same value. This serves as a useful check!

For you, when t = 1.776 s, x_{you} = 4.73 m.

For the Porsche, when t = 1.776 s, x_{Porsche} = 4.73 m.