An object with cross section A is shot horizontally across frictionless ice. Its initial velocity is v₀ₓ at t₀ = 0 s. Air resistance is not negligible. a. Show that the velocity at time t is given by the expression vₓ = v₀ₓ --------------- 1 + C𝓭pAv₀ₓt / 2m

Video transcript

Hey, everyone. So this problem is working with drag forces. Let's see what they're giving us. We have an athlete of mass m ice skating on a frictionless surface. On a windy day, the athlete stops pushing at time T zero when she reaches a horizontal velocity of V sub A. So she continues to slide in the horizontal direction while the drag force decreases her speed, assume that the cross sectional area, the athlete is a and were asked to derive the expression of the horizontal velocity as a function of time. So the first thing we're gonna do here is draw our free body diagram. So we know that because it is a frictionless surface And she stops pushing right at time T zero. That the only force acting in the X direction is our drag force and that's going to be acting in the negative horizontal direction. And then of course, we have her weight and the normal force from Newton's second law, we can recall that the sum of the forces and this time we're gonna be working in the X direction, right are horizontal so that some of the forces equals mass times acceleration in the horizontal direction, the only force we have is this drag force again acting in the negative. And we can recall that our drag force Is given by 1/2. There's a row C A B squared. We can also recall that our acceleration is simply the derivative of velocity with respect to time. And so A X is equal to D V X D T. So when we substitute those into our first equation, we come up with one half rho C A V squared, the X squared, we're working in the X direction equals M D V X D T. So let's rearrange this so that we have our V X and D V X on the same side of the equation. And that gives us row C A B row C A D T over to um equals negative. So V X squared or one over V X squared. And we take that to the other side, I'm gonna rewrite that as V X to the negative two D V X. Alright. So from there, we can take the integral of both sides. We'll pull out the constants here. So row C A to M are all constants on this side. Integral from T not T 0 to T D T equals negative. The integral of the X D T D V X. We'll take that from B A which is our initial velocity giving it the problem to be X. When we take those into girls. We are left with, grow, see A over to em times T from Tina to T equals one over the X from B A to B X. We will solve that out to get row C A over to em T minus T, not Equals one over the X -1 over the a. Now they're asking us for the horizontal velocity as a function of time. So it's going to be V X. So we want to isolate V X here, we can do that looks like the X equals to um B A over row C A T minus T, not times V A plus two M. You can simplify that just one step further. We're left with V A over row C A T minus T, not times B A that over to em plus one. And so that is our answer how we have our horizontal velocity as a function of time. And so then when we go to look at our potential answer choices that aligns with choice and so A is the correct answer here. And so that's all we have for this problem. We'll see you in the next video.