OpenStax College Physics, Chapter 8, Problem 13 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. A 75 kilogram person is riding in a car with an initial velocity of 20 meters per second and they collide with some sort of concrete thing on a bridge, I guess a bridge abutment, and they hit the dashboard in part A which compresses one centimeter, which is one times ten to the minus two meters, and we're asked to figure out what force will they experience given this amount of distance over which they're stopped. Then in part B we'll replace the one centimeter compression of the dashboard padding with a 15 centimeter compression of the airbag and compare the force in that case. So the net force that they experience is mass time acceleration and the sideways force exerted by the dashboard will be the net force because it's only going to be the one horizontal force on the person. So the work here involves finding acceleration. So from kinematics we know that the final velocity squared is initial velocity squared plus two times acceleration times the distance and the final velocity is zero. So we can subtract v i squared from both sides and then switch the sides around and we have two a d equals negative v i squared. Then divide both sides by two d and you get the acceleration is negative initial velocity squared divided by two times the distance over which they are brought to a stop. Then we substitute that into our F net equals m a formula by replacing a with all this here. So we have F net is negative m v initial squared over two d. Thus negative 75 kilograms times 20 meters per second squared all divided by two times one times ten to the minus two meters. This is negative 1.5 times ten to the six newtons. The force is negative because it's in the opposite direction to the initial velocity which we took to be positive. Then we replace the dashboard padding with an airbag which has a 15 centimeter compression. In that case the force works out to being an order of magnitude less, it's negative 1.00 times ten to the five newtons.