surviving-a-large-deceleration-on-july-131977-while-on-a-test-drive-at-britains-silverstone-racetrac

Question

Surviving a Large Deceleration On July $13,1977,$ while on a
test drive at Britain's Silverstone racetrack, the throttle on David
Purley's car stuck wide open. The resulting crash subjected Purley
to the greatest $^{4} g-$ force" ever survived by a human-he decelerated
from 173 $\mathrm{km} / \mathrm{h}$ to zero in a distance of only about 0.66 $\mathrm{m} .$ Calculate the magnitude of the acceleration experienced by Purley
(assuming it to be constant), and express your answer in units of
the acceleration of gravity, $g=9.81 \mathrm{m} / \mathrm{s}^{2} .$

Nathan Silvano · Accepted Answer

So in this problem we have experiment in this experiment. Ah, human survive a deceleration when he grows from a velocity Initial velocity off 173 limiters per hour to a final velocity off zero filament To spur our, uh in just in a distance off general point 66 meters. We want to calculate the gravity&#x27;s acceleration experienced by the humor. Okay, so let&#x27;s see have a variation of velocity and I have a distance or we can use be scratch equals vizier Scratch less to a doubt x So here we&#x27;re going to find that the final velocity zero the initial velocity 17 tree Actually, let&#x27;s change the kilometers per hour, two meters per second and the way which we do This is just dividing one, 73 for 3.6. So this is going to be equal two 3.6 is great to be equal to 48 point 05 meters per second. So let&#x27;s calculate that we have here 48 point 05 square. Well, us too. Hey, times 0.66 So Okay, let&#x27;s calculate this. The acceleration that the human experience was just was equal to one 1000 749 point five meters per second squared. If we divided this by Dexter The Gravity deceleration We&#x27;re going to have the acceleration experience by the human deceleration because to one seven eight point tree tree Jeez, rich Jesus, the gravity, the separation. So that&#x27;s your final answer. Thanks for watching.

Averell Hause · Answer

so here the even the initial velocity of the car is 100 kilometers per hour. This would be multiplied by one meter per second for every 3.6 kilometers per hour. And this is equaling 27.8 meters per second of a meters per second. Approximately, we&#x27;re going to choose Max initial to be zero, of course. And that is going to be the point at which the deceleration begins again. We haven&#x27;t accelerate a deceleration of negative 30 g and what they mean is simply negative. 30 times 9.8 meters per second squared. Ah, this is equaling Approximately two negative 294 meters per second squared and we can find the displacement using equation to 12 C s. We can say that to philosophy Final squared again. We don&#x27;t have the time. So if we don&#x27;t have the time, we need to use the to Deacon O Matic to t k O Matic equation that does not have a time term. So the final squared equals the initial squared plus two times a doubt the X and in this case, we can solve for Delta X, so Delta X would be equal. Uh, we&#x27;re coming to a stop. So velocity final squared would be zero. So it would be negative. The initial squared divided by two times a day, and this would be equal to negative 27 0.8 meters per second quantity squared, divided by two times negative. 294 meters per second squared. And we find that the Auto X is gonna equal 1.31 meters. That is, uh, the final answer. That is the end of the solution. Thank you for watching.

Bryce Samwel · Answer

Okay, what&#x27;s actually start by doing part B? Because I don&#x27;t actually help us get part A. So what&#x27;s know here that this is part B we&#x27;re starting with. So acceleration is equal to the change in velocity over the change in time. So that&#x27;s pretty straightforward in this problem. In this problem, the change in velocity our final velocity is 27 0.3, minus our initial velocity, which is 17.4 in the change in time. Well, the total time elapsed is 10 seconds. And when we plugged that into our calculator, we find that our acceleration 0.99 meters per second squared. That&#x27;s pretty straightforward and it will actually help us get party. Let&#x27;s use that acceleration for part A. There&#x27;s a couple different equations you could use here. But for the sake of this problem, let&#x27;s use this one where the final velocity squared is equal to the initial velocity squared, plus two times the acceleration times the distance. Now we know the final and initial velocities as well as the acceleration, so that should help us get X. And when we solve for X algebraic lee first you would subtract over the initial velocity. And then once you had two. X by itself, who just divide by two A as we do here and we get this equation. So it&#x27;s plug in the numbers we have and find the value of X. So once again, 27.3. But it&#x27;s squared minus 17.4. Squared all over to times are acceleration, which was 0.9 nine. And that gets us a final answer of 223 25 when we played that into our calculator.

Emily Reynolds · Answer

for this question, we are analyzing the situation of a seat belt and a car stopping, and we first have to start off with some unit conversions. So it gives us 30 G&#x27;s that we have to convert into meters per second squared. Okay, so it tells us that one G is 9.8. So 30 G&#x27;s times 9.8 meters per second squared is one G. So we find that 30 G&#x27;s is equal Teoh 294 meters per second squared. Then we also have to convert our initial velocity before the car stops, which is 100 kilometers per hour. We have to convert that into meters per second. So in one kilometer there&#x27;s 1000 meters and then in one hour there&#x27;s 3600 seconds. So converting our initial velocity, we find that it is 27.78 meters per second. Okay, we also know that the final velocity is zero meters per second cause they&#x27;re coming to rest. Um, again, the acceleration we converted, that&#x27;s gonna be the 294 meters per second squared. And then we are ultimately trying to figure out the displacement or how much the front end of the car has to crumple, um, to allow this to happen. So the formula I want to use is the X squared equals V I squared plus two a Delta X again The final is gonna be zero 27.78 squared was two times to 94 times are Delta X And then when we solve for Delta X, we find that it is 1.31 meters.

Ryan Hood · Answer

The first thing to note about this problem is that I made a negative instead of positive because it opposes the motion that you have originally this be not here. Another thing. The notices that the units on the V not do not agree with the units on the A. So I&#x27;m going to convert the units on the scene on Sophie not is one hundred and five cloners per hour. I&#x27;m going to multiply this by a conversion factor. First go and change the kilometers and the meters now in one kilometer there&#x27;s a thousand meters. So that&#x27;s that conversion and the kilometers cancels. I also need to convert the hours. So ours goes up here in the seconds one hour, there&#x27;s thirty six hundred seconds and then the hours cancel. The result of this is twenty nine point two meters per second and then we can actually proceed with problem. In this problem, you want to calculate X my sex, not we want this and we&#x27;re going to calculate using the threshold acceleration so that we know a threshold distance needed. And then if we go below that, we know we&#x27;re in trouble. We have to go above the special distance, but it&#x27;s nice to have the threshold distance and what we&#x27;re going to do that calculate it is used the chemical equation Musical Veena on squared plus two a x my sex Not I use this one because we have no information about the time and this is the one can a magic equation that doesn&#x27;t involve time. I&#x27;m gonna calculate on the next page just shifting things around to solve for X my sex nine I get this expression No. And then now I&#x27;m gonna plug in values that we have so x my sex nine is equal to zero. Since the final velocity after the airbag has stopped us a zero and the initial velocity which of the initial velocity squared which is a hundred fifty two point six four over two times acceleration Negative. Two fifty. This ends up being one point seven one meters and so if you move less than one point seven one meters, then you&#x27;re in trouble. But if you&#x27;re moving over one point seven one or one point seven of one meter staring this interaction, then you&#x27;re gonna be okay. And that&#x27;s the conclusion of this problem. That&#x27;s the answer

Zachary Warner · Answer

We&#x27;re told that the maximum acceleration that the human body can experience and survive is 250 meters per second. And we&#x27;re in a car wreck here where the initial speed of the car is vey not and it&#x27;s evil. 105 kilometers per hour and I put that into meters per second by using the fact that there are 1000 meters per kilometer and there are 3600 seconds per hour. This comes out to be 29.17 meters per second and the final velocity V zero meters per second. Since we end up rest were asked to calculate the distance that this airbag must stop you at an order for this acceleration to take place. And you survived the crash. So by definition of our Kin O Matic equation, here we have v squared. The final velocity squared is equal to the initial velocity squared plus two times acceleration. But we&#x27;re de accelerating here, right? We&#x27;re not accelerating or d accelerating. So this is gonna be negative, eh? So a b e either accelerating or d accelerating free, and you can you can&#x27;t survive an acceleration greater than 250 or a de acceleration. And here we&#x27;re stopping. So it&#x27;s negative. And then this is multiplied by the distance over which this occurs. The distance. Here is what we&#x27;re trying to solve for so we&#x27;ll solve this equation for distance V squared to zero. So we&#x27;ll let this go to zero. What we have here is the distance is equal to negative V not squared up, divided by two times. Negative, eh? So, plucking the values of this accretion in front. The distance is equal to 1.70 meters. We can go ahead and box set in as a solution to the question.

Problem

A boat is cruising in a straight line at a consta…

Problem 58 Medium Difficulty

Answer

178.3$g$

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