A car is traveling around a horizontal circular track with radius r = 210 m as s

Question

A car is traveling around a horizontal circular track with radius r = 210 m as shown. It takes the car t = 60 s to go around the track once. The angle ?A = 19° above the x axis, and the angle ?B = 64° 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 pont B?

Explanation / Answer

Here ,

for the angular speed of car , w = 2 pi/T

w = 6.282/60

w = 0.105 rad/s

1)

acceleraion of car = w^2 * r

acceleraion of car = 0.105^2 * 210

acceleraion of car = 2.302 m/s^2

the acceleraion of car is 2.302 m/s^2

2)

at point A ,

VAx = r * w * (-cos(thetaA))

VAx = 210 * 0.1047 * (-cos(19))

VAx = -20.8 m/s

3)

VAY = r * w * (sin(thetaA))

VAy = 210 * 0.1047 * sin(19)

vAy = 7.16 m/s

4)

ABx = r * w^2 * cos(thetaB)

ABx = 210 * 0.1047 * cos(64)

ABx = 1 m/s^2

the x component of the car's acceleration when it is at point B is 1 m/s^2

5)

ABy = 210 * 0.1047^2 * sin(64)

ABy = 2.07 m/s^2

the y - compoenent of car's acceleration at B is 2.07 m/s^2

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