A car is traveling around a horizontal circular track with radius r = 180 m as s

Question

A car is traveling around a horizontal circular track with radius r = 180 m as shown. It takes the car t = 67 s to go around the track once. The angle 9A = 21® above the x axis, and the angle 9g = 59: below the x axis. 1) What is the magnitude of the car's acceleration? 1.58 m/s^2 2) What is the x component of the car's velocity when it is at point A m/s | Submit 3) What is the y component of the car's velocity when it is at point A m/s | Submit 4) What is the x component of the car's acceleration when it is at point B m/s- 1 Submit, 5) What is the y component of the car's acceleration when it is at point B m/s- | Submit, W 6) As the car passes point B. the y component of its velocity is increasing constant decreasing

Explanation / Answer

1)   speed = 2 *3.14* 180/67

                   = 16.871 m/sec

     => acceleration = 16.871 *16.871/180

                                  = 1.58 m/sec2

2)   x component of velocity at point a = - 16.871 * cos 69

                                                              = - 6.046 m/sec

3)   y component of velocity at point a = 16.871 * sin 69

                                                              = 15.75 m/sec

4) x component of acceleration at point B = - 1.58 * cos59

                                                                     = - 0.8137 m/sec2

5)   y component of acceleration at point B = 1.58 * sin59

                                                                     =   1.354 m/sec2

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

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