Physics - proving the stopping distance of a car

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SUMMARY

The discussion focuses on calculating the stopping distance of a car traveling at 120 km/h with a reaction time of 0.20 seconds and a deceleration of 8.0 m/s². The correct stopping distance is determined to be 76 meters, which requires considering both the distance traveled during the reaction time and the distance during braking. The initial misunderstanding involved incorrectly assuming the car stops within the reaction time, rather than accounting for the time before braking begins.

PREREQUISITES
  • Understanding of kinematic equations, specifically SUVAT equations.
  • Knowledge of unit conversion from km/h to m/s.
  • Familiarity with concepts of acceleration and deceleration.
  • Basic grasp of physics principles related to motion and stopping distances.
NEXT STEPS
  • Study the SUVAT equations in detail to understand their applications in motion problems.
  • Learn how to convert units between kilometers per hour and meters per second effectively.
  • Research the effects of reaction time on stopping distances in various driving scenarios.
  • Explore real-world applications of braking distance calculations in automotive safety.
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and safety measures related to stopping distances.

totomyl
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Homework Statement


an accident occurs up ahead on the highway. a driver traveling at 120km/h [e] reacts in 0.20s and applies the brakes causing an acceleration of 8.0m/s2 [w]. show that the stopping distance is 76 m.
what am i doing wrong? i changed the acceleration to match the directions, so i made it negative. but i am not getting the right answer.

Homework Equations


d = vi * t + 0.5(a * t^2)

The Attempt at a Solution


i attempted this by using:

d = (120km/h / 3.6[e])(0.20s) + 0.5(-8.0m/s[e] * 0.20s^2)

d = 6.5m?
 
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totomyl said:

Homework Statement


an accident occurs up ahead on the highway. a driver traveling at 120km/h [e] reacts in 0.20s and applies the brakes causing an acceleration of 8.0m/s2 [w]. show that the stopping distance is 76 m.
what am i doing wrong? i changed the acceleration to match the directions, so i made it negative. but i am not getting the right answer.

Homework Equations


d = vi * t + 0.5(a * t^2)

The Attempt at a Solution


i attempted this by using:

d = (120km/h / 3.6[e])(0.20s) + 0.5(-8.0m/s[e] * 0.20s^2)

d = 6.5m?
The problem is you have assumed that the car goes from 120 kph to 0 kph in 0.2 s, which is not what the problem states. The driver takes 0.2 s to press the brake pedal after he sees the accident ahead of him.

You should pick another SUVAT equation which relates distance, acceleration,and initial and final velocity.
 
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SteamKing said:
The problem is you have assumed that the car goes from 120 kph to 0 kph in 0.2 s, which is not what the problem states. The driver takes 0.2 s to press the brake pedal after he sees the accident ahead of him.

You should pick another SUVAT equation which relates distance, acceleration,and initial and final velocity.
So, I just have another question, in this question the final velocity would be at a stop, so 0 m/s correct? and also would i have to find the distance traveled before pressing on the brakes and add it to the distance it took while slowing down?
 
totomyl said:
So, I just have another question, in this question the final velocity would be at a stop, so 0 m/s correct? and also would i have to find the distance traveled before pressing on the brakes and add it to the distance it took while slowing down?
Yes and yes.
 
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SteamKing said:
Yes and yes.
thank you, your answers have been very helpful and i am very grateful!
 

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