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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

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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