Surviving a Car Crash with Airbag Deployment: Calculating Stopping Distance

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Homework Help Overview

The discussion revolves around calculating stopping distances in two different scenarios involving vehicles: one involving an airbag deployment during a crash and the other concerning a car's decision-making at a traffic light. The subject areas include kinematics and dynamics related to motion and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to determine stopping distance after a crash, questioning the correctness of their calculations. Other participants suggest converting units and using alternative equations for efficiency. A new scenario is introduced regarding a car's motion at an intersection, prompting requests for general problem-solving strategies.

Discussion Status

The discussion is ongoing, with participants exploring different methods and equations to approach the problems. Some guidance has been offered regarding unit conversion and alternative equations, but no consensus has been reached on the best approach for either scenario.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance provided. There is also an emphasis on understanding the implications of acceleration and stopping distances in real-world scenarios.

courtrigrad
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The human body can survive a negative acceleration trauma incident if the magnitude of the acceleration is less than 250 m/s^2. If you are in an automobile accident at an initial speed of 96 km/h and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?


So I know that [itex]v_{0} = 96[/itex], [itex]v_{x} = 0[/itex] and [itex]a_{x} = 250[/itex]. So is it correct to say [itex]v_{x} = v_{x}_{0} + a_{x}t[/itex] to find the time, or [itex]0 = 96-250t[/itex] and [itex]t = 0.384 sec[/itex]? Then you use [itex]x-x_{0} = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}[/itex] and you get the distance to be [itex]18.432 m[/itex]

Is this correct?

Thanks
 
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any ideas?
 
change 96 km/h to m/s.
 
I would do the above suggestion and use this equation... it's faster.

[tex]V_{f}^2=V_{0}^2+2ad[/tex]
 
A car 3.5 m in length and traveling at a constant speed of 20 m/s is approaching an intersection. The width of the intersection is 20 m. The light turns yellow when the front of the car is 50 m from the beginning of the intersection. If the driver steps on the brake, the car will slow at -4.2 m/s^2. If the driver instead steps on the gas pedal, the car will accelerate at 1.5 m/s^2. The light will be yellow for 3.0 s. Ignore the reaction time of the driver. To avoid being in the intersection while the light is red, should the driver hit the brake pedal or gas pedal?

Could someone give me a general idea of where to start, and a general problem solving strategy?

Thanks
 

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