Seat belts and air bags save lives by reducing the forces exer...

Question

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a 'crumple zone' in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm. A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Accepted Answer

Hey, everyone in this problem, a crash test dummy equipped with a force sensor is used to measure the impact force on a passenger during a frontal collision. 55 kg dummy is buckled up with a seatbelt and the car initially moving at a speed of 60 kilometers per hour collides with a thick wall. The airbag is deployed and brings the dummy to rest at a distance of 125 centimeters or asked to determine the magnitude of the impact force on the dummy from the airbag as measured by the force sensor. We have four answer choices here all in Newton's option, a 306 option B 540 option C 6111 and option D 66,000. So we're looking for a force here, we know the mass of this dummy. So let's recall that a force is equivalent to the mass multiplied by the acceleration. And that's Newton's second law. Now again, we know the mass but we don't know the acceleration. Ok. So the first thing we need to do is calculate the acceleration and then we can come back to our force calculation. So let's think about the dynamics of this crash and think about what we know and we know that the initial speed that the car is traveling is 60 kilometers per hour. Now, we want to convert this into our standard unit of meters per second because our answer choices are in newtons. Ok. So we need our acceleration to be in that meters per second squared unit. So that when we work out our force, it will be in newtons, ok? All right. So we're gonna multiply, we're gonna multiply by 1000 m per kilometer, ok? The unit of kilometers will divide out. We're gonna be left with meters, ok? And then we're gonna multiply by one hour divided by 3600 seconds, ok? Because there are 3600 seconds in every hour. The unit of hour is gonna divide out and now we're left with meters per second exactly like we wanted. And this value is gonna be 16.66 repeated meters per second, ok? So this is the initial speed that the car is traveling at and we know that the final speed is going to be zero, ok? This car is brought to rest and the dummy is brought to rest. More importantly here, we're talking about the dummy here, the dummy is brought to rest and so our final speed VF is going to be 0 m per second. Now we're told this happens in a distance of 125 centimeters. And so delta X distance traveled is going to be 125 centimeters. Again, converting to our standard unit. We want this in meters, we divide by 100 we get 1.25 m. And what we're looking for right now is the acceleration and we don't have any information about the time. So we're gonna choose a kinematic or U AM equation with these four variables. And we're gonna see what we get. Now, the equation that we're gonna use VF squared is equal to V not squared plus two, multiplied by a multiplied by delta X substituting in our values 0 m per second squared is equal to 16.66 repeated meters per second. All squared plus two multiplied by a multiplied by 1.25 m. Yeah, on the left hand side, we just have zero we want to solve for a. So we're gonna move this term with the A to the left hand side and simplify a little bit. OK. So we're gonna subtract from both sides and we get that negative 2.5 m multiplied by A is equal to 277.77 repeated meters squared per second squared A R 2.5 came from the two multiplied by the 1.25. And then we squared our V knot to get that value on the right hand side of our equation. Now, we can divide both sides by negative 2.5. And we get that our acceleration A is going to be negative 111.11 m per second squared. All right. And this makes sense. We got a negative value and that makes sense because our crash dummy is decelerating, they're slowing down, they're coming to rest. So negative acceleration makes sense. Now we're gonna work out the force. OK? We needed the acceleration first. We've got that so we can get back to our force calculation. And recall again, Newton's second law tells us that that is the mass multiplied by the acceleration. In this case, that's gonna be 55 kg multiplied by negative 111.11 m per second squared, which gives us a force. I love about negative 6111.11 repeated newtons, right? All right. Now, this question was asking for the magnitude of the impact force. And so we're just gonna take the absolute value of the force we found. So we get 6111.11 repeated news. And that is that impact force on the crash dummy by the airbag. OK. And again, we are told in the problem that the airbag is what brings the dummy to rest. So we're assuming that all of that force is from the airbag. Ok. Let's compare this to the answer choices that we were given. We can see that our answer corresponds with option C 6111 noons. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Newton's First Law of Motion

Crumple Zones

Force and Deceleration

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