A car is traveling at 120 km/h (75 mph).When applied the braking system can stop the car with a deceleration rate of 9.0 m/s².The typical reaction time for an alert driver is 0.5 s versus 2 s for a sleepy driver.Assuming a typical car length of 5 m,calculate the number of additional car lengths it takes the sleepy driver to stop compared to the alert driver. A)13 B)3.0 C)10 D)16 E)26

Question

Quizplus · Accepted Answer

First, we need to convert the initial velocity from km/h to m/s:

120 km/h = 33.3 m/s

Next, we can use the following formulas to calculate the stopping distance for each driver:

Alert driver: 
d = (v^2)/(2a) + L 
where v = initial velocity = 33.3 m/s 
a = deceleration rate = 9.0 m/s^2 
L = car length = 5 m

d = (33.3^2)/(2*9.0) + 5 
d = 117.8 m

Sleepy driver: 
d = (v^2)/(2a) + L 
but with a longer reaction time (2 s instead of 0.5 s), so the car travels an additional distance before braking: 
d = v*t + (v^2)/(2a) + L 
where t = 2 s

d = 33.3*2 + (33.3^2)/(2*9.0) + 5 
d = 156.2 m

The difference between the stopping distances is:

156.2 m - 117.8 m = 38.4 m

To find the number of additional car lengths, we divide this difference by the length of one car:

38.4 m / 5 m = 7.68

So the sleepy driver takes approximately 8 additional car lengths to stop compared to the alert driver. The closest choice to this value is C) 10.

A Car Is Traveling at 120 Km/h (75 Mph)