Answer
The point of no return is 3260 feet from the start of the runway.
The time available to decide on a course of action is $25.5$ seconds.
Work Step by Step
We can convert the runway distance to units of feet:
$1.50~miles \times \frac{5280~ft}{1~mi} = 7920~ft$
Let $x$ be the point of no return:
$v_f^2 = v_0^2 + 2ax$
$v_f^2 = 0 + (2)(10.0)x$
$v_f^2 = 20.0~x$
Note that $v_f$ from the previous part of the question is the initial velocity when the plane starts decelerating. We can find an expression for $v_f^2$ when the plane is decelerating at a rate of $7.00~ft/s^2$:
$0 = v_f^2+(2)(-7.00)(7920-x)$
$v_f^2=(14.0)(7920-x)$
We can equate the two expressions for $v_f^2$ and solve for $x$:
$20.0~x = (14.0)(7920-x)$
$34.0~x = 110,880$
$x = 3260~ft$
The point of no return is 3260 feet from the start of the runway.
We can find the time it takes to reach this point:
$x = \frac{1}{2}at^2$
$t = \sqrt{\frac{2x}{a}}$
$t = \sqrt{\frac{(2)(3260~ft)}{10.0~ft/s^2}}$
$t = 25.5~s$
The time available to decide on a course of action is $25.5$ seconds.
