Define general expression for the instantaneous acceleration

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Q. For no apparent reason, a poodle is running at a constant speed of 6.00 m/s in a circle with radius 2.9 m. Let v1 be the velocity vector at time t1 , and let V2 be the velocity vector at time t2 . Consider (change in v)=v2-v1 and (change in t)=t2-t1. Recall that.
A-ave = v/t

For t = 0.4 s compute the direction (relative to v1 ) of the average acceleration A-ave.

For t = 0.1 s compute the magnitude (to four significant figures) of the average acceleration A-ave.

For t = 0.1 s compute the direction (relative to v1) of the average acceleration A-ave .

For t = 6×10-2 s compute the magnitude (to four significant figures) of the average acceleration A-ave.

state your answer using four significant figures.

For = 6×10-2 compute the direction (relative to v1 ) of the average acceleration A-ave.

estimate your results to the general expression for the instantaneous acceleration a for uniform circular motion that is derived in the text.

The magnitude of a-ave is less than of a and a-ave tends to a as t is decreased.

The magnitude of a-ave is greater than of a and a-ave tends to as t is decreased.

Reference no: EM1387351

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