Question

" A car is travelling around a horizontal circular track with radius r = 230 m at a constant speed v = 17 m/s as shown. The angle θA = 25° above the x-axis, and the angle θB = 58° below the x-axis.

1. What is the x component of the car’s acceleration when it is at point A?
2.What is the y component of the car’s acceleration when it is at point A?
3.What is the x component of the car’s acceleration when it is at point B?
4. What is the y component of the car’s acceleration when it is at point B?"

Answer 1

r = 230 m
v = 17 m/s

Formula: |Ac| = v^2 / r

Point A

|Ac| = v^2 / r = (17m/s)^2 / (230m) = 1.257 m/s^2

Components:

cos(25) = - [ x-component / |Ac| ] => x-component = -|Ac|*cos(25)
sin(25) = - [y-component / |Ac| ] => y-component = - |Ac|*sin(25)

x-component = - 1.256 * cos(25) m/s^2 = - 1.139 m/s^2
y-component = -1.256 * sin (25) m/s^2 = - 0.531 m/s^2

Point B

|Ac| = 1.256

x-component = - 1.256 cos(58) = - 0.666 m/s^2
y-component = + 1.256 sin(58) = + 1.065 m/s^2