Apparently an easy Dynamics question

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The discussion revolves around a physics problem involving a sports car losing contact with the road during a high-speed chase. Participants analyze how to determine the radius of curvature at the crest of a hill and the force exerted on the driver as the car passes through that point. Key equations discussed include centripetal acceleration and the relationship between forces acting on the driver. Conversions between units, such as miles per hour to feet per second, are emphasized for accurate calculations. The conversation highlights the complexity of solving the problem without a specific equation for the hill's profile.
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During a high-speed chase, a 2400-lb sports car traveling
at a speed of 100 mi/h just loses contact with the road as it
reaches the crest A of a hill. a) Determine the radius of
curvature r of the vertical profile of the road at A. b)
Using the value of r found in a part a), determine the
force exerted on a 160-lb driver by the seat of his 3100-lb
car as the car, traveling at a constant speed of 50 mi/h,
passes through A.



Homework Equations


an= v^2/p, p is what we are searching from, but it if we are not given an equation of the hill, it seems unlikely that it can be solved.
p can also be found using derivatives of the position function...


Is this even solve-able?
 

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hi there, fluffypuffin! welcome to pf! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
fluffypuffin said:
an= v^2/p, p is what we are searching from

contact is lost when the reaction force is zero …

so do F = ma for the driver, using your centripetal acceleration formula …

what do you get? :smile:
 
You'd have to make some conversions because velocity is in seconds. So
100 mi/h= 146.6 ft/s then
p= 146.6 ^2/ 32.2

Correct?

Thank you so much!
 
looks good! :biggrin:

(are you ok on the other parts?)
 
Nope! Want to help me? :)
 
if we sum the forces in the normal direction
wouldn't it be:
mg- N= m (50)^2/ 668.04

??
 
fluffypuffin said:
mg- N= m (50)^2/ 668.04

yes :smile: (except you need to convert the 50 into ft/s :wink:)
 
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