##### Homework: Relative And Circular Motion 3

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A car is traveling around a horizontal circular track with radius r = 230 m as shown. It takes the car t = 63 s to go around the track once. The angle θA = 27° above the x axis, and the angle θB = 59° below the x axis.

5)

What is the y component of the car’s acceleration when it is at point B
Sep 14th, 2015

velocity= distance/time

distance is circumference

Circumference = 2 * π * radius

=2 *π  * 230

= 460 * π

v = 460 * π ÷ 63

= 22.9386 m/s

cars acceleration = v^2/r

= 22.9386 ^2 / 230

= 2.2877 m/s^2

To determine the y component of the velocity at point B

acceleration vector points toward the origin, so subtact from 180,  180º -59

y = a * sin θ
y =  2.2877 * sin (180 -59)

= 1.9609

= 1.96  m/s^2

y component of the car’s acceleration when it is at point B = 1.96  m/s^2

Please let me know if you need any clarification. I'm always happy to answer your questions.....thank you
Sep 14th, 2015

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Sep 14th, 2015
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Sep 14th, 2015
Oct 21st, 2017
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