Physics - proving the stopping distance of a car

AI Thread Summary
To calculate 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, it is crucial to account for the distance traveled during the reaction time before braking begins. The initial speed must be converted to meters per second, and the stopping distance should be calculated using the appropriate SUVAT equations that incorporate both the distance covered during the reaction time and the distance required to come to a stop. The final velocity is indeed 0 m/s when the car stops. The correct approach involves summing the distance traveled during the reaction time with the distance covered while decelerating. This method will yield the accurate stopping distance of 76 meters.
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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