A car is traveling around a horizontal circular track with radius r = 270 m as s

Question

A car is traveling around a horizontal circular track with radius r = 270 m as shown. It takes the car t = 67 s to go around the track once. The angle ?A = 20 above the x axis, and the angle ?B = 57 below the x axis.

1) What is the magnitude of the cars acceleration?

2) What is the x component of the cars velocity when it is at point A

3) What is the y component of the cars velocity when it is at point A

4) What is the x component of the cars acceleration when it is at point B

5) What is the y component of the cars acceleration when it is at point B

6) As the car passes point B, the y component of its velocity is

increasing

constant

decreasing?

Explanation / Answer

1.acceleration a= v^2/r= ((2*3.14*270)/67)^2/270 =2.372 m/s^2

2.velocity =(2*3.14*270)/67 =25.307 m/s

x component at A =Vsin 20=25.307*sin 20=8.6555 m/s

3.y component at A = Vcos 20=25.307*cos 20=23.7808m/s

4.a=2.372 m/s^2,

x compeonent at B = a cos 57 = 1.292 m/s^2

5.y compeonent at B = a sin 57 = 1.989 m/s^2

as theta is decreasing, hence cos theta must be increasing ..hence velocity is increasing

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