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Physics question involving go-karts?
A flat go-kart track consists of two straightways 100 m long with semicircular ends which have a radius of 25.0 m. A go-kart executes the curved end with a constant tangential acceleration while slowing from 13.8 m/s to 5.6 m/s. Determine the magnitude of the total acceleration of the go-kart at the beginning and end of the curve. (Answers should be to two decimal places.)
2 Answers
- NCSLv 76 years agoFavorite Answer
tangential acceleration:
arc length s = πr
v² = u² + 2as = u² + 2πar
(5.6m/s)² = (13.8m/s)² + 2π*a*25.0m
a = -1.01 m/s² ◄ tangential acceleration, constant through turn
Beginning of turn:
centripetal a = v²/r = (13.8m/s)² / 25.0m = 7.62 m/s²
total a = √(cent a² + tang a²) = 7.69 m/s² ◄ total, beginning of curve
End of turn:
cent a = (5.6m/s)² / 25.0m = 1.25 m/s²
total a = √(cent a² + tang a²) = 1.61 m/s² ◄ total, end of curve
Hope this helps!
