A person who is properly restrained by an over-the-shoulder se...

Question

A person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 'g's' (1.00 g = 9.80 m/s². Assuming uniform deceleration at 30 g's, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 95 km/h.

Accepted Answer

Hey, everyone in this problem, we're asked to imagine a scenario where a high speed train is traveling at a velocity of 160 kilometers per hour along a straight track upon detecting an obstacle. The train's emergency braking system is activated, subjecting it to a uniform deceleration of two GS and where one G is equal to 9.8 m per second squared. We're asked to calculate the distance required for the train safety systems to engage and bring it to a complete stop ensuring passenger safety during that deceleration. So we have four answer choices all in meters. Option A 44.4 option B 65.3 option C 50.4 and option D 11.1. So we have a kinematics or U AM type problem here and we have a velocity that we're given. We know that we want this to come to a complete stop. We're given some information about acceleration. So let's write out all of these variables that we're told. Ok. So the initial velocity VN is 160 kilometers per hour and that's the speed the train is traveling before they start using those breaks. Now the answer choices are in meters, ok? The acceleration is gonna be in meters per second squared. So let's convert this velocity into meters per second. Our standard unit. Ok. So we're gonna multiply by 1000 m per kilometer, ok? And then we're gonna multiply again by one hour per 3600 seconds, ok? And doing this is gonna allow the unit of hours the unit of kilometers to vi divide out and we're gonna be left with meters per second. And while we work it out, it's gonna be 44.44 repeated meters per second. OK. So that's our initial velocity. In that standard unit, our final velocity, it's gonna be just 0 m per second. OK? We wanna find the distance required for this to come to a complete stop. So that final velocity is gonna be zero not moving. Now, our acceleration we're told is negative two GS. So negative two multiplied by G which is going to be and why it's negative is because it's a deceleration. OK. So the magnitude is two G but it's a deceleration, it's slowing the train down. So we have that negative in front of it. OK. So negative two multiplied by 9.8 m per second squared, which is gonna be negative 19.6 m per second squared. OK. All right. And delta X we'll call it the distance travel is what we want to find. OK? Now we have three known values VNV FA and unknown that we're looking for delta X. We can just use the kinematic or U AM equation with these four variables substitute in what we know and solve that equation is going to be VF squared is equal to V, not squared plus two multiplied by a multiplied by delta X. OK. So we have an equation without t we don't have any information about the time. T that's not what we're looking for. And so we're using this equation now substituting in our values. VF is gonna be 0 m per second, all squared. That's going to equal the knot, which we found to be 40 4.44 repeated meters per second. That's all squared plus two multiplied by negative 19.6 m per second squared multiplied by delta X. Now we're gonna take this second term, this two multiplied by negative 19.6 multiplied by DE X. We're gonna move it to the left hand side by adding it on both sides. And we get that 39.2 meters per second squared, multiplied by delta X is equal to 44 144 repeated meters per second, all squid. And the final step is to divide both sides by 39.2 m per second squared. OK. On the right, we have the units of meters squared per second squared. We're dividing by meters per second squared. So we're gonna be left with just a unit of meters, which is what we want for this problem. So that's good. Everything checks out and we're gonna get that delta X is equal to 50.39 m approximately. So comparing to our answer choices, rounding to one decimal place, we could see that the correct answer here is going to be option see 50.4 m. Ok? It takes 50.4 m for this train to come to a complete stop. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Deceleration and G-Forces

Kinematic Equations

Energy Absorption in Crumple Zones

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