AP Physics please helpp me????

Question

An object, after being released from its circular path, travels the distance OA in the same time it would have moved from O to P on the circle. The speed of the object on and off the circle remains constant at the same value. Suppose that the radius of the circle is 4.0 m and the angle is is 29°. What is the distance OA?
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Answer 1

If it travelled at the same speed, it covers the same distance, so the question is really asking for the arc length it would have covered.

Angle = arc length/radius (but angle is in radians, so you must convert).

Answer 2

This simply asks for how far on the circle (c) the object would have traveled and then straighten that distance out (c = S) to find out how far it did travel at "speed" |v|, which is the magnitude of the velocity. In which case c = S = |v|t; where t is the time of travel at v. This follows because distance traveled (D) is speed (|v|) times time (t) traveled no matter what the path of travel is.

So the object travels some portion of the circle circumference (C = 2pi R); where R = 4 meter. "Some portion" means a ratio, c/C = (theta/360) of C; where theta = 29 degrees.

Thus, the portion the object would have traveled c = (29/360) C = (29/360) 2 pi R; where R = 4 meters. And c = S = |v|t; where |v| and t are the same for the curved path and the straight path. Thus, your straight path OA = S = c = (29/360) 2 pi R and you can do the math.

The physics in this is that v is the tangential velocity around the circle C, but it is the linear velocity along the straight path S. Even so, as the speeds (magnitudes of the velocities) are the same, their directions have no bearing on the distance traveled. In your case c = S = |v|t (where |v| is speed having no direction) is the key.

Answer 3

minimum on the right of the arc, optimal on the backside (the two ends) of the arc. Acceleration through gravity is the reason. through fact the ball is going up, its velocity decreases through strain of gravity until its efficient velocity is 0 (vertically). because it is going backtrack it good points acceleration until it contacts the floor. offered the landing element is on the comparable height through fact the commencing element, the two ends might have the comparable (opposite) velocity. Horizontal velocity remains consistent by using the test as long as we are discounting air resistance, hence its result on the relative velocity is omitted.