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 calculate the direction (relative to v1 ) of the average acceleration A-ave.
For t = 0.1 s calculate the magnitude (to four significant figures) of the average acceleration A-ave.
For t = 0.1 s calculate the direction (relative to v1) of the average acceleration A-ave .
For t = 6×10-2 s calculate the magnitude (to four significant figures) of the average acceleration A-ave.
Express your answer using four significant figures.
For = 6×10-2 calculate the direction (relative to v1 ) of the average acceleration A-ave.
Compare 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.