Question

Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the brakes than a sober driver's car? Assume that both are initially traveling at 50.0 mi/h and their cars have the same acceleration while slowing down, and that the sober driver takes 0.33 s to hit the brakes in a crisis, while the drunk driver takes 1.0 s to do so. (5280 ft = 1 mi)

Answer 1

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

ΔS = 49.13 ft