## College Physics (4th Edition)

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.
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.