The car is traveling at a constant speed through a dip in the road. The radius of curvature of the road at point A, the bottom of the dip, is 500 m. What speed of the car, measured in km/h, would result in an acceleration of magnitude 0.2g, where g is the gravity, when the car is at A? 500 m A

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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The subject is all about dynamics. The topic is all about the Curvilinear Motion of Particles or Tangential and Normal Components. Please use two decimal places and show the conversion process. Use pen and paper to solve. Thank you. 

Here are the formulas: 

a_x=0  a_y=-g

  v_x=v_xo  v_y=v_yo-gt

  x=x_o+v_xo t  y=y_o+v_yo t-1/2 gt^2

for the Tangential and Normal Components fomulas:

a_t=dv/dt                a_n=v^2/ρ

 

The car is traveling at a constant speed through a dip in the road. The radius of curvature of the
road at point A, the bottom of the dip, is 500 m. What speed of the car, measured in km/h, would
result in an acceleration of magnitude 0.2g, where g is the gravity, when the car is at A?
500 m
A
Transcribed Image Text:The car is traveling at a constant speed through a dip in the road. The radius of curvature of the road at point A, the bottom of the dip, is 500 m. What speed of the car, measured in km/h, would result in an acceleration of magnitude 0.2g, where g is the gravity, when the car is at A? 500 m A
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