Question 388744
Split this into two parts: the time that the car is going at a constant speed, and the time that the car is decelerating.


With the driver's reaction time 1 second, the car travels 16 m since it is going at 16 m/s.


After, the car decelerates with an acceleration of -4 m/s^2. Since v = at + v_0,


{{{0 m/s = (-4 (m/s^2))t + 16 m/s}}} --> t = 4 seconds. Using the equation


{{{X = (1/2)at^2 + (v_0)t + x_0}}} we plug in values a = -4 m/s^2, t = 4 s, v_0 = 16 m/s, x_0 = 16 m (since the car already traveled 16 m)


{{{X = (1/2)(-4 (m/s^2))(4 s)^2 + (16 m/s)(4 s) + 16 m}}}


{{{X = 48 m}}}