Question

Alcohol consumption slows people's reaction times. In a controlled government test, it takes a certain driver t1=0.39 s to hit the brakes in a crisis when unimpaired and t2=0.87 s when drunk. If the car is initially traveling at v=95 km/h, how much farther does the car travel before the brakes are applied when the person is drunk than it travels when the person is sober?

Answer 1

Answer:

Extra distance traveled is 12.6 m

Explanation:

Initial constant speed of the car is given as

$v = 95 km/h$

so it is given as

$v = 95 \times \frac{1000 m}{3600 s}= 26.4 m/s$

now we know that the reaction time is 0.39 s when driver is unimpaired

so distance covered is given as

$d_1 = 26.4 \times 0.39 = 10.3 m$

when driver is drunk then reaction time is 0.87 so the distance covered is given as

$d_2 = 26.4 \times 0.87 = 23 m$

so the extra distance traveled by the person when he is drunk is given as

$d = d_2 - d_1$

$d = 23 - 10.3$

$d = 12.6 m$

Answer 2

We have equation of motion $S = ut + \frac{1}{2} at^2$, here S = displacement, u = initial velocity, t is the time taken and a is the acceleration.

Time taken is the difference in time in the response times = 0.87 - 0.39 = 0.48 seconds

Initial velocity = 95 km/hour = 95*5/18 = 25 m/s

Acceleration = $0 m/s^2$

Substituting $S = 25*0.48+\frac{1}{2} *0*0.48^2 = 12 m$

So the car travel 12 m before the brakes are applied when the person is drunk than it travels when the person is sober.