In this problem, We're gonna talk about centripetal acceleration. Eso consider a particle. That's when they're going certain emotion as shown here, the bishop and the motion as the radius R Uh, and we know that the particle is being accelerated with a contributor acceleration, Alfa, we can decompose the total acceleration of this particle in two components. The first one is a radio components. So suppose that the party is here. The radio component called this interpretive exploration We'll be pointing towards the center off rotation, and it's equal to V squared over R where V is the linear speed off the particle or omega squared times are in this accounts for the change in direction off the particle and exists even when there is no change in speed. Okay, uh, the tangential exploration, on the other hand, accounts for only for the change in speed. So it's the derivative of the speed with respect to time and can be written as the angular acceleration times. The radios are okay. So in our problem, we have a car that is in a circular path of radius 23.5 m and at a certain point in the path the car is still accelerating and has a new angular speed of 0.5 71 radiance per second. And we have the information that the angle that the total speed does with the radius. So suppose that the car is right here on then the total Not I had big before, but it should be total exploration. This angle here is equal to 23.5 degrees. I'm sorry. It's actually 35 degrees. Sorry for the mistake on our goal is to find what is the value off the total exploration, the magnitude of the tour exploration. Well, notice that we have that the acceleration can be decomposed into two components. So we have one component That is the centripetal acceleration that points towards the center of rotation. And the other one that Is that the general exploration? Yes. So this is this attributable. This is the financial, and this is the total one. Uh, actually, I wrote it as a C. But it should be a t. Um, So notice that this angle here is theta. So a T is equal to a C divided by the cold sign of theta a c. The centripetal acceleration is omega squared times are and then we have to divide it by coastline of data. So 80 is equal to 0.5 71 radiance per second squared times are and that is 23.5 m and we have to divide it by the coal sine of the angle theta, which is 35 degrees mhm. Eso 80 is equal to 9.35 m per second square. So this is a total exploration and the answer to your question, Yeah, what?