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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 θA = 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

Respuesta :

Centripetal acceleration of car is given by formula

[tex]a_c = \frac{v^2}{R}[/tex]

now plug in the values in this

[tex]a_c = \frac{23^2}{210}[/tex]

[tex]a_c = 2.52 m/s^2[/tex]

Part b)

At position A we have

x component of acceleration is given as

[tex]a_x = -a_c cos23[/tex]

[tex]a_x = -2.32 m/s^2[/tex]

Part c)

At position A we have

y component of acceleration is given as

[tex]a_y = -a_c sin23[/tex]

[tex]a_y = -0.98 m/s^2[/tex]

Part d)

At position B we have

x component of acceleration is given as

[tex]a_x = -a_c cos53[/tex]

[tex]a_x = -1.52 m/s^2[/tex]

Part e)

At position B we have

Y component of acceleration is given as

[tex]a_y = a_c sin53[/tex]

[tex]a_y = 2.01 m/s^2[/tex]

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