It takes a 615-kg race car 14.3 s to travel at a uniform speed around a circular racetrack of 50.0 m radius .
What is the acceleration of the car?
What average force must the track exert on the tires to produce this acceration?
2007-11-13 12:09:38 · 2 answers · asked by james 2 in Science & Mathematics ➔ Physics
The speed around a circular racetrack is
V=2piR/t
1. Centripital acceleration is
a=V^2/R
a= (2piR/t)^2 /R
a= R((2pi/t)^2
a=50(( 2 pi / 14.3)^2=
a=9.65 m/s^2
2. Centripetal force is
F=ma
F=615 x 9.65=
F=5930 N
2007-11-13 12:17:26 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
A=(4[pi]^2r)/t^2
A=(4[pi]^2(50))/14.3^2
A=1973.1762/14.3^2
A= 9.649 m/s/s
F=MA
F=615 x 9.649
F= 5934.135 N
You may want to check my math. The calculator on my computer is hard to use
2007-11-13 12:23:42 · answer #2 · answered by wil 3 · 0⤊ 1⤋