Alert driver vs Sleepy Driver distance traveled

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SUMMARY

The discussion focuses on calculating the additional distance traveled by a sleepy driver compared to an alert driver when braking from a speed of 120 km/h (75 mph) with a deceleration of 9.0 m/s². The alert driver has a reaction time of 0.50 seconds, while the sleepy driver has a reaction time of 2.0 seconds. The calculations reveal that the sleepy driver travels an additional 10 car lengths (50 meters) before stopping. The initial step using the formula v² = Vo² + 2a(Δx) was deemed unnecessary for this specific problem, as a simpler approach using d = vt suffices.

PREREQUISITES
  • Understanding of basic kinematics, including distance, speed, and acceleration.
  • Familiarity with the equations of motion, specifically v² = Vo² + 2a(Δx) and d = vt.
  • Knowledge of reaction times and their impact on stopping distances.
  • Ability to perform unit conversions between kilometers per hour and meters per second.
NEXT STEPS
  • Study the implications of reaction time on driving safety and accident prevention.
  • Learn more about the physics of braking and stopping distances in vehicles.
  • Explore advanced kinematic equations for varying acceleration scenarios.
  • Investigate the effects of different deceleration rates on stopping distances in various vehicle types.
USEFUL FOR

This discussion is beneficial for driving instructors, automotive safety researchers, and anyone interested in understanding the dynamics of vehicle stopping distances and the effects of driver alertness on safety.

smsport
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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 0.50 s versus 2.0 s for a sleepy driver. Assuming a typical car length of 5.0 m, calculate the number of additional car lengths it takes the sleepy driver to stop compared to the alert driver.

I worked this out first by using v^2=Vo^2 + 2a(Δx) to get the distance traveled
0= 33.3m/s^2 + (-9m/s^2)(x)
x= 61.7m

Then, I used d= vt for driver #1: (33.3m/s)(.5s)= 16.7m
and for driver #2: (33.3m/s)(2.0s)= 66.7m

Then, I added 61.7m to each driver's distance:
driver #1 total dist traveled=78.4m
driver #2 total dist traveled= 128.4m

Then, subtract 78.4m from 128.4m to get the total extra distance traveled by the sleepy driver. Then, divide this figure by 5m to get total car lengths additional.

128.4m- 78.4m = 50m/5m= 10 car lengths.

Now, after going through the problem it appears my initial step is really unnecessary because I could have just used d=vt for driver #1 and driver #2 and took the difference of those figures divided by 5m to get car lengths extra for the sleepy driver. Am I correct? Unless the question was asking for total distance traveled by both or either driver, then step #1 isn't even needed and this problem could have solved much quicker.

Please confirm.

Thanks!
 
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Yes

d1=vt1 +(v2/2a)

d2=vt2 +(v2/2a)

Δd=d2-d1
 
Last edited:

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