## Physics: Principles with Applications (7th Edition)

$v_0 = (95 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 26.4 ~m/s$ We can use the velocity $v_0$ to find the collapse distance $\Delta y$ of the car's front end: $\Delta y = \frac{v^2 - v_0^2}{2a}$ $\Delta y = \frac{0 - (26.4 ~m/s)^2}{(2)(30)(9.80 ~m/s^2)}$ $\Delta y = 1.19 ~m$ If the collapse distance of the car's front end is 1.19 meters, the person should not feel a deceleration of more than 30 g's.