Let the distance from A to P be \( x \). Since BP is thrice of AP, the distance from P to B is \( 3x \). As given, the speed of Car 2 is half of Car 1, let's denote the speed of Car 1 as \( v \), and thus the speed of Car 2 is \( \frac{v}{2} \).
Let \( t \) be the time taken by Car 1 to reach P from A. Therefore, Car 1 covers the distance \( x \) in time \( t \), so:
\[ x = v \cdot t \]
The time taken by Car 2 to reach P from B is \( t + 1 \) hours, as Car 2 arrives 1 hour after Car 1. Car 2 covers the distance \( 3x \) in \( t + 1 \) hours, so:
\[ 3x = \frac{v}{2} \cdot (t + 1) \]
Substituting \( x = v \cdot t \) into the second equation gives:
\[ 3(v \cdot t) = \frac{v}{2} \cdot (t + 1) \]
Canceling out \( v \) from both sides and solving for \( t \):
\[ 3t = \frac{t + 1}{2} \]
Multiplying through by 2 to clear the fraction:
\[ 6t = t + 1 \]
Simplifying this equation results in:
\[ 6t - t = 1 \]
\[ 5t = 1 \]
\[ t = \frac{1}{5} \] hours.
To convert the time from hours to minutes, multiply by 60:
\[ \frac{1}{5} \times 60 = 12 \text{ minutes} \]
Thus, the time taken by Car 1 to reach point P from A is 12 minutes, which falls precisely within the expected range of \( 12, 12 \).