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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

Respuesta :

Refer to the diagram shown below.

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

Answer: 1.96 m/s²

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