College Algebra Exit Project

ryan_kalle

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Part B: Speed Limit
You are asked to find a speed limit for a certain stretch of road on Pioneer Trail going south towards New Smyrna Beach. The road is used primarily by teenagers as short cut to go into town. There have been many accidents because of speeding and the city is asking their Daytona State College intern to find the right speed limit so that accidents will be reduced to a minimum. In other words, the city wants to post a speed limit sign on this stretch of road. You’re taking your newly designed car for a test drive out on the road. The stretch of road consist of a curve not unlike a semi circle, see picture.
As a Traffic engineering student at Daytona State, you are trying to find a comfortable radius for the curve, and the right distance needed to stop the car. You must take into account the fact that a curves’ radius, must be large enough for the car not to leave the road in either way. The Department of Transportation provides the following equation for the minimum radius and the distance needed for a car to stop. The minimum radius (in feet) that should be used is given by r= 0.334s^2 and the distance, d, needed to stop a car is given by d= 0.05 s^2 + 1.1 s, where s is the expected speed of traffic in miles per hour. Please answer the following questions: (Round your answers to the nearest tenths)

a. If the expected speed of your newly designed car is 20 mph, 40 mph, what should the minimum radius of the curve be using the formula r= 0.334s^2?

r(s)=.334s^2

r(20)=.334(20)^2
r(20)=133.6

r(40)=.334(40)^2
r(40)=534.4

b. Since we doubled the speed in a, does this automatically mean the radius doubled as well? What is the relationship between the speed and the radius?

No, the necessary radius increased by a multiple of 4.

c. How fast can you go around the curve with a radius of 500 feet?

s(r)=sqrt(r/.3340

s(500)=sqrt(500/.334)
s(500)=38.7

d. Use your answer from step (c) to find the distance needed to stop the car using the formula
d= 0.05 s^s + 1.1 s?

d=.05(38.7)^2 + 1.1(38.7)
d=117.5
e. What speed limit should be posted on the road where the drivers drive around the curve and have 80 feet to come to a stop?

Section e in this part
.05s^2 + 1.1s = 80
It looks like a quadratic to me. How would I go about solving this?
 
ryan_kalle said:
Section e in this part
.05s^2 + 1.1s = 80
It looks like a quadratic to me. How would I go about solving this?
First, multiply by 100 to get rid of decimals:
5s^2 + 110s - 8000 = 0
Divide by 5:
s^2 + 22s - 1600 = 0

Now solve that using the quadratic formula.
 
Denis said:
ryan_kalle said:
Section e in this part
.05s^2 + 1.1s = 80
It looks like a quadratic to me. How would I go about solving this?
First, multiply by 100 to get rid of decimals:
5s^2 + 110s - 8000 = 0
Divide by 5:
s^2 + 22s - 1600 = 0

Now solve that using the quadratic formula.

I need a face palm emoticon.
 
Subhotosh Khan said:
Denis said:
[quote="ryan_kalle":1cx67q24]I need a face palm emoticon. <<< What is a face palm emoticon ??? Just in case I need one - I need to know when I found one!!!
[/quote:1cx67q24]
It's like when you pulled a "duhhh...", then slap your forehead...kapish?
In your case, Subhotosh, that would be quite often :wink:
 
Denis said:
Subhotosh Khan said:
[quote="ryan_kalle":1i6p3mvd]I need a face palm emoticon. <<< What is a face palm emoticon ??? Just in case I need one - I need to know when I found one!!!
It's like when you pulled a "duhhh...", then slap your forehead...kapish?
In your case, Subhotosh, that would be quite often :wink: <<< I refuse to comment on this hockey-puck-chaser's rude comment :mrgreen: :twisted: :mrgreen:
[/quote:1i6p3mvd]

Oh you mean this one..... :oops:

or

this one..... :eek:
or


this one..... :eek: :idea:
 
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