Answer
The minimum stopping distance for 60 mi/h is $~48~m$
Work Step by Step
Let the magnitude of deceleration be $a$. We can find an expression for the stopping distance $x_1$ when the initial speed is $v$:
$v_f^2 = v^2+2(-a)x_1$
$x_1 = \frac{0-v^2}{2(-a)}$
$x_1 = \frac{v^2}{2a}$
Let the magnitude of deceleration be $a$. We can find an expression for the stopping distance $x_2$ when the initial speed is $2v$:
$v_f^2 = (2v)^2+2(-a)x_2$
$x_2 = \frac{0-(2v)^2}{2(-a)}$
$x_2 = 4\times \frac{v^2}{2a}$
$x_2 = 4~x_1$
Since $x_1 = 12~m$, the minimum stopping distance for 60 mi/h is $~48~m$
