What is the minimum stopping distance for a car traveling at a speed of 40 m/s?

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SUMMARY

The minimum stopping distance for a car traveling at a speed of 40 m/s is determined to be 100 meters, factoring in a driver's reaction time of 0.5 seconds. The calculations involve using the equations of motion, specifically Vf = Vi + at and d = ((Vi + Vf)/2) * t. A user initially calculated the stopping distance as 96.7 meters but later corrected their approach by applying the equation v² = u² + 2as to find the correct acceleration and stopping distance. This discussion highlights the importance of accurately applying kinematic equations in physics problems.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of acceleration and its calculation
  • Familiarity with the concept of reaction time in driving scenarios
  • Ability to interpret position-versus-time graphs
NEXT STEPS
  • Study the derivation and application of kinematic equations
  • Learn about the effects of reaction time on stopping distances
  • Explore graphical representations of motion, specifically position-versus-time graphs
  • Investigate real-world factors affecting vehicle stopping distances, such as road conditions and vehicle weight
USEFUL FOR

Students studying physics, automotive engineers, driving instructors, and anyone interested in understanding vehicle dynamics and safety measures.

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[SOLVED] Minimum Stopping Distance

Homework Statement


The minimum stopping distance for a car traveling at a speed of 30m/s is 60m, including the distance traveled during the driver's reaction time of 0.50s.

a.What is the minimum stopping distance for the same car traveling at a speed of 40 m/s?

b. Draw a position-versus-time graph for the motion of the car in part a. Assume the car is at Xo=0m when the driver first sees the emergency situation ahead that calls for a rapid halt.

Homework Equations


For a part:
Vf=Vi+at

d=((Vi+Vf)/2)*t

The Attempt at a Solution



The correct answer for a part is 100m. I'm getting 96.7m. I started with the given and figured the time to be 5.335s, using Vf=Vi+at, and subtracted .5s for the reaction time. Then I used d=((Vi+Vf)/2)*t, which I got 96.7m. I don't think that I'm plugging into the formulas correctly?
 
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The car will not stop in the same time. But its acceleration is constant. Try [itex]v^2 = u^2 +2as[/itex]
 
Okay, I got the acceleration from the given information and then used the equation you have given to solve for a. part. Thanks for the help--it is greatly appreaciated.
 

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