Is the Angle Between Acceleration and Velocity Vector Greater Than 90 Degrees?

[KatlynEdwards]

Homework Statement

An object moves clockwise with decreasing speed around an oval track. There are velocity vectors at two points (G and H). G has a greater speed than H. Is the angle between the acceleration and velocity vector of G greater than, less than, or equal to 90 degrees?

Homework Equations

N/A

The Attempt at a Solution

So I know when it's in constant motion the acceleration vector is perpendicular to the velocity and point's towards the inside of the oval. But since in this instance it's slowing down, that means that the delta v vector is angled because one of the legs of the triangle is shorter. So my thoughts are that it is larger than 90 degrees. Is this the right line of thinking?

Thanks for your input!

 

[PhanthomJay]

Your answer may be correct but your reasoning is not. The problem is asking for the angle between the velocity vector (not the change in velocity vector) and acceleration vector at a point G. The velocity vector is tangent to the curve at that point, and points in the direction of the tangential motion. One component of the acceleration vector points inward toward the center of the curvature (centripetal acceleration), and the other acceleration component (at a right angle to the centripetal acceleration vector) is tangent to the curve, pointing opposite to the direction of the velocity vector because it is slowing down. Draw a quick sketch showing the velocity vector and the resultant of the centripetal and tangential acceleartion vectors (the total or net acceleration) and determine if the angle betwen the two is >, <, or = to 90 degrees.

 

[KatlynEdwards]

Oh, I was under the impression that the acceleration vector was parallel to the change in velocity vector, is this false?

 

[PhanthomJay]

Well that is true...but the velocity vector is changing due to both its change in direction and its change in magnitude...its change in direction gives it a centripetal acceleration, v^2/r, pointing inward to the center of curvature, and it's change in speed gives it a tangential deceleration...the vector addition of those 2 acceleration vectors is the direction of the acceleration...so at point G, the velocity vector is tangent to the curve, but the acceleration vector is not...the angle between the 2 is greater than, less than, or equal to 90 degrees? (Your answer is correct...I am not sure from your wording though if you understood why...).

 

[rwisz]

Yes, your line of thinking is correct. In this scenario, the acceleration vector will be pointing towards the center of the oval track, while the velocity vector will be tangent to the track. Since the object is slowing down, the angle between these two vectors will be greater than 90 degrees. This can also be seen by considering the definition of acceleration as the rate of change of velocity, which means the acceleration vector will be in the opposite direction of the change in velocity vector. Therefore, the angle between the two vectors will be greater than 90 degrees.