24. The constant-acceleration condition permits the use of Table 2-1. (a) Setting v= 0 and x0= 0 in 22002()vvaxx=+−, we find 26201411(5.0010)0.100 m .221.2510vxa×=−=−×Since the muon is slowing, the initial velocity and the acceleration must have opposite signs.(b) Below are the time-plots of the position
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