A car is traveling around a horizontal circular track with radius r = 220 m as shown. It takes the car t = 61 s to go around the track once. The angle θA = 23° above the x axis, and the angle θB = 57° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s velocity when it is at point A
3) What is the y component of the car’s velocity when it is at point A
4) What is the x component of the car’s acceleration when it is at point B
5) What is the y component of the car’s acceleration when it is at point B
Because the time to go around the track is 61 s, the angular velocity is ω = (2π rad)/(61 s) = 0.103 rad/s The tangential velocity is v = rω = (220 m)*(0.103 rad/s) = 22.66 m/s
Part 1) The centripetal acceleration is a = v²/r = (22.66 m/s)²/(220 m) = 2.334 m/s² The centripetal acceleration is directed toward the center of the circle.
Answer: 2.334 m/s²
Part B)
The x-component of the velocity at A is - v*sin(23°) = -22.66*sin(23°) = - 8.854 m/s
Answer: - 8.854 m/s
Part 3) The y-component of the velocity at A is v* cos(23°) = 22.66*cos(23°) = 20.86 m/s
Answer: 20.86 m/s
Part 4) The x-component of the acceleration at B is - a*cos(57°) = - (2.334 m/s²)*cos(57°) = -1.27 m/s²
Answer: - 1.27 m/s²
Part 5) The y-component of acceleration at B is a*sin(57°) = (2.334 m/s²)*sin(57°) = 1.96 m/s²
