Answer
(a) The magnitude of the driver's acceleration is $420.5~m/s^2$
(b) The magnitude of the passenger's acceleration is $4205~m/s^2$
Work Step by Step
(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 $4205~m/s^2$
