How Does Decreasing Speed Affect the Angle of Delta-v in Oval Track Motion?

[WarpDrive]

"An object moves clockwise with decreasing speed around an oval track." There is also a drawing with two velocity vectors drawn. I was able to draw the vector "delta"-v, based on the two velocity vectors, but how does it change as one point moves closer to the other? In UCM with constant velocity, the angle formed by delta-v between two points and the velocity vector at one of the points approaches 90 degrees. How does that change if the object has decreasing velocity? I was thinking that delta-v might tend towards some angle greater than 90 degrees, as part of it must point towards the center, and part of it must point in the opposite direction of the velocity vector to decrease the speed of the object. I think that that angle might approach some angle less than 90 degrees if the object has increasing velocity. However, I'm not sure about my arguments, so is this how it is, or am I missing something?

Thanks!

 

[WarpDrive]

Sorry, I forgot to read the errata concerning this problem. How does the change in speed of an object affect the angle between the acceleration vector and the velocity vector at a point. In UCM, the acceleration vector is perpendicular to the velocity vector. If the object's speed is increasing, an additional tangential component of acceleration causes the acceleration vector to point at less than 90 degrees. If the object's speed is decreasing, in this case, I would think the angle would be greater than 90 degrees. But does any of this change because I'm dealing with an oval?

Thanks!