Asked by kg90
A 1310-kg racecar is driving around a circular track of radius 153 m. In the instant shown in the figure, the vehicle has a forward speed of 31.7 m/s and is slowing at a rate of 5.35 m/s2. What is the magnitude of the net force acting on the vehicle at this time and the force\'s direction relative to the velocity?
Answers
Answered by
Elena
Normal (or centripetal) acceleration is
a(n) = v²/R = 31.7²/153=6.57 m/s²
Tangential acceleration
a(τ)= -5.35 m/s²
Acceleration
a=sqrt{ a(n)²+ (5.35)²} =8.47 m/s²
tanφ=5.35/6.57=0.814
φ=31.2º
α=90+φ=90 º +31.2 º =121.2 º
a(n) = v²/R = 31.7²/153=6.57 m/s²
Tangential acceleration
a(τ)= -5.35 m/s²
Acceleration
a=sqrt{ a(n)²+ (5.35)²} =8.47 m/s²
tanφ=5.35/6.57=0.814
φ=31.2º
α=90+φ=90 º +31.2 º =121.2 º