## College Physics (4th Edition)

The minimum stopping distance for 60 mi/h is $~48~m$
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 $~48~m$