Respuesta :
Answer:
The steady deceleration, [tex]a = -27.57 m/s^{2}[/tex]
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[tex]\times 0.45 = 40.5 m/s[/tex]
reaction time of motorist, t = 0.5 s
Solution:
Distance traveled before applying brakes, s' = v't
s' = [tex]40.5\times 0.5 = 20.25 m/s[/tex]
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:
[tex]v^{2} - u^{2} = 2as[/tex]
where
v = final velocity
u = initial velocity
s = distance traveled
a = acceleration
Now,
[tex]v^{2} - u^{2} = 2as[/tex]
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:
[tex]0 - 40.5^{2} = -2a\times 29.75[/tex]
[tex] 1640.25 = - 59.5a[/tex]
[tex]a = -27.57 m/s^{2}[/tex]
where negative sign is indicative of deceleration
