## College Physics (4th Edition)

(a) The magnitude of the driver's acceleration is $420.5~m/s^2$ (b) The magnitude of the passenger's acceleration is $4205~m/s^2$
(a) We can find the acceleration of the driver: $v_f^2 = v_0^2+2ax$ $a = \frac{v_f^2-v_0^2}{2x}$ $a = \frac{0-(29~m/s)^2}{(2)(1.0~m)}$ $a = -420.5~m/s^2$ The magnitude of the driver's acceleration is $420.5~m/s^2$ (b) We can find the acceleration of the passenger: $v_f^2 = v_0^2+2ax$ $a = \frac{v_f^2-v_0^2}{2x}$ $a = \frac{0-(29~m/s)^2}{(2)(0.100~m)}$ $a = -4205~m/s^2$ The magnitude of the passenger's acceleration is $4205~m/s^2$