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OleMash [197]
3 years ago
14

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

A = 23° above the x axis, and the angle θB = 53° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s acceleration when it is at point A
3) What is the y component of the car’s acceleration 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
Physics
1 answer:
Fantom [35]3 years ago
6 0

Centripetal acceleration of car is given by formula

a_c = \frac{v^2}{R}

now plug in the values in this

a_c = \frac{23^2}{210}

a_c = 2.52 m/s^2

Part b)

At position A we have

x component of acceleration is given as

a_x = -a_c cos23

a_x = -2.32 m/s^2

Part c)

At position A we have

y component of acceleration is given as

a_y = -a_c sin23

a_y = -0.98 m/s^2

Part d)

At position B we have

x component of acceleration is given as

a_x = -a_c cos53

a_x = -1.52 m/s^2

Part e)

At position B we have

Y component of acceleration is given as

a_y = a_c sin53

a_y = 2.01 m/s^2

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