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 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?
1 answer