Answer:
The steady deceleration,
![a = -27.57 m/s^(2)](https://img.qammunity.org/2020/formulas/physics/college/buz2wlmmvlcwealahdvo8kppdsp4ic8a0u.png)
Given:
1 mile = 0.45 m/s
initial distance of the motorist, s = 50 cm
speed of the motorist at the time he noticed the traffic light, v = 90 mi/h = 90
![* 0.45 = 40.5 m/s](https://img.qammunity.org/2020/formulas/physics/college/ppoog593q53x3ebv1wzr744zas90hgxquk.png)
reaction time of motorist, t = 0.5 s
Solution:
Distance traveled before applying brakes, s' = v't
s' =
![40.5* 0.5 = 20.25 m/s](https://img.qammunity.org/2020/formulas/physics/college/l04p2pwk6i3vgboq863dn8819xf5882pov.png)
The remaining distance traveled to the traffic light, s'' = 50 - s' = 50 - 20.25 = 29.75 m
Now, using third equation of motion to calculate steady deceleration:
where
v = final velocity
u = initial velocity
s = distance traveled
a = acceleration
Now,
![v^(2) - u^(2) = 2as](https://img.qammunity.org/2020/formulas/physics/college/d86wr2aou4zj8hhcaissqllt1nxvw0ops9.png)
In this eqn w.r.t the given case v = 0, as brakes are applied and the body will decelerate as a result of braking and u = v' and s = s'' = 29.75 m
Now, eqn becomes:
![0 - 40.5^(2) = -2a* 29.75](https://img.qammunity.org/2020/formulas/physics/college/n7bkylguq2d89r9hfccvlxsnz5wyq8ys4k.png)
![1640.25 = - 59.5a](https://img.qammunity.org/2020/formulas/physics/college/kp5qdjn8snqadhser1yecogg582w0hzej0.png)
![a = -27.57 m/s^(2)](https://img.qammunity.org/2020/formulas/physics/college/buz2wlmmvlcwealahdvo8kppdsp4ic8a0u.png)
where negative sign is indicative of deceleration