Asked by Daniel Limon
A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 "'s."
What distance is traveled if the person is brought to rest at this rate from 115 ?
What distance is traveled if the person is brought to rest at this rate from 115 ?
Answers
Answered by
Henry
a = 30gs = 30 * 9.8m/s^2 = -294 m/s^2.
Vo=115km/h*1000m/km*1h/3600s=31.94m/s.
V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a.
d = (0-1020.4)/-588 = 1.74 m.
Vo=115km/h*1000m/km*1h/3600s=31.94m/s.
V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a.
d = (0-1020.4)/-588 = 1.74 m.
Answered by
Ink
IF YOU ARE HAVING HARD TIME UNDERSTAND HENRY'S ANSWER READ THIS:
To simplify what Henry said if you need clarification
30g's is 30 times the acceleration of the earth within the first 10000 feet of the atmosphere, so in this case -294m/s^2 is the acceleration
To find the distance convert your velocity to m/s and divide that by 2 times your acceleration
Distance = (velocity^2)/-588)
To simplify what Henry said if you need clarification
30g's is 30 times the acceleration of the earth within the first 10000 feet of the atmosphere, so in this case -294m/s^2 is the acceleration
To find the distance convert your velocity to m/s and divide that by 2 times your acceleration
Distance = (velocity^2)/-588)
Answered by
bahog igit
kaon tae
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.