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 ?
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)
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.
To calculate the distance traveled during deceleration, we can make use of the formula:
distance = (initial velocity)^2 / (2 * acceleration)
In this case, the initial velocity is given as 115 mph, and we need to convert it to an appropriate unit of measurement like feet per second (fps) because acceleration is typically given in fps^2.
First, let's convert the initial velocity from miles per hour (mph) to feet per second (fps). We can use the conversion factor of 1 mph = 1.47 fps.
So, the initial velocity in fps is:
initial velocity = 115 mph * 1.47 fps/mph ≈ 169.05 fps
Next, we need to find the acceleration. Given that the deceleration is no more than 30 "s" (which I assume means 30 feet per second squared), we can directly use this value in the formula:
acceleration = -30 fps^2 (negative sign represents deceleration)
Now we can plug the values into the formula to find the distance traveled:
distance = (initial velocity)^2 / (2 * acceleration)
= (169.05 fps)^2 / (2 * -30 fps^2)
= 28588.4025 fps^2 / -60 fps^2
≈ -476.47 feet
Since distance cannot be negative in this context, we take its absolute value to get the positive distance.
Therefore, the distance traveled when brought to rest at a deceleration of 30 fps^2 from an initial velocity of 115 mph is approximately 476.47 feet.