Calculating Minimum Stopping Distance for a Car at Different Speeds

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SUMMARY

The minimum stopping distance for a car traveling at 38.0 m/s is calculated by first determining the distance traveled during the driver's reaction time of 0.560 seconds, which is 21.28 m. The deceleration, calculated using the equation v² - u² = 2as, is 10.416 m/s². The total stopping distance combines the reaction distance and the braking distance, resulting in a final stopping distance of 90.58 m for the car at 38.0 m/s.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² - u² = 2as
  • Knowledge of basic physics concepts such as acceleration and deceleration
  • Familiarity with calculating distance based on speed and time
  • Ability to perform unit conversions and basic arithmetic operations
NEXT STEPS
  • Learn about kinematic equations in detail, focusing on their applications in real-world scenarios
  • Research the effects of different reaction times on stopping distances
  • Explore advanced topics in vehicle dynamics, including braking systems and their efficiencies
  • Study the impact of speed on stopping distances in various weather conditions
USEFUL FOR

Physics students, automotive engineers, driving instructors, and anyone interested in understanding vehicle stopping distances and safety measures.

klm
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The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the drivers reaction time of 0.560 s.
what is the min. stopping distance for the same car traveling at a speed of 38.0 m/s?

tried to use this equation v^2 - u^2 = 2 a s
but then didnt know what to do.
 
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Use that equation to figure out the car's acceleration. What distance did you use? (How far has the car traveled before the brakes are applied?)
 
43.2m is the real distance he first travels with the velocity of 30 m/s. because i subtracted 16.8 m from the 60 m.
so my a= 10.416
 
Good! Now apply that--in reverse--to the second case, where the speed is 38 m/s, to find the new total stopping distance. (You'll have to calculate a new "reaction time distance".) The car's acceleration remains the same.
 
i don't know how to find a new reaction time distance??
i used this formula :
0^2 - 38^2 = 2(10.416) x
and found x to be 69.3m but this is incorrect. can you please help me.
 
klm said:
i don't know how to find a new reaction time distance??
Do it the same way as before. How far does the car move before the brakes are applied?
 
38 x .56?
=21.28 m
but what do i do with that?
 
You add it to the other distance. Understand what happens when the person wants to stop: (1) there's a time delay due to reaction time, so the car moves some distance before the brakes are even applied; (2) once the brakes are applied, the car accelerates to a stop.

The total stopping distance is the sum of both of these distances.
 
you add 21.28 to 60? to get 81.28m?
 
  • #10
klm said:
you add 21.28 to 60? to get 81.28m?
No. 60m was the total stopping distance for the 30m/s case--nothing to do with the 38 m/s case!

Add 21.28 m to the distance you calculated for the de-accelerating phase of the motion.
 
  • #11
to the 69.3?
69.3+21.28 = 90.58?
sorry if i leave really quick i have to run to class in about 2 mins!
 
  • #12
Yes.
 
  • #13
THANK YOU! so so much
 

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