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How to find the time for a curved path ? 
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#1
Dec2212, 08:24 AM

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1. The problem statement, all variables and given/known data
Al right so i am given this diagram. They are asking about the time for the whole journey. The car is travelling on a steep hill through the section of AD and which continues on to DC with a small curved path.Find the time for this journey. The car starts from rest at point A. 3. The attempt at a solution I used this equation to find the distance from A to D a^2= b^2 +c^2 Then i used this equation to find the time from A to D V2  V1 = 2gd to find V2 Then i used the equation v2 = v1 + gt to find the time T don't know how to account for the curved path and how should i continue? Here is a diagram: http://s1176.beta.photobucket.com/us...3ea8a.jpg.html 


#2
Dec2212, 08:49 AM

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P: 1,053

A "small curved path"? How small is small? Without any constraints on smallness, you could make it as small as you want. Like nonexistently small. Or maybe the size of the curve of the tire of a car. Either way, not significant.



#3
Dec2212, 10:09 AM

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#4
Dec2212, 12:11 PM

P: 17

How to find the time for a curved path ?
V2^2  V1^2 = 2*g*d You can confirm with dimensional analysis. Also at the end there, you're only getting the time from the last segment, not the whole trip. You still need to calculate the time for the first segment unless you used your third equation twice 


#5
Dec2212, 03:51 PM

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#6
Dec2212, 04:01 PM

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P: 11,890

For DC, where do you see an acceleration? I would calculate the velocity with energy conservation (or simply use the value calculated at AD)



#7
Dec2212, 04:07 PM

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#8
Dec2212, 04:15 PM

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P: 11,890

If you don't know about energy conservation, don't use it. You can calculate the velocity with other formulas as well.



#9
Dec2212, 04:17 PM

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#10
Dec2212, 04:28 PM

P: 17

The only acceleration is due to gravity when the car is going down the ramp.. As soon as it is on the straight section, the velocity is constant (no acceleration)



#11
Dec2212, 04:56 PM

P: 217

And do i need to break the gravity in half..... where they will become... sinG and cosG ??? thats when you do tension... i don't think so right.. because its just time or do you? 


#12
Dec2212, 07:05 PM

P: 17

You will need to break acceleration into the components acting in the x direction. As you mentioned, you found the length of the slant (AD)
therefore ax(acceleration in the X direction) is g*COS(theta) = g*(BD/AD) then you can use kinematics to solve time. Xf=xo+Vot+1/2at^2 Vf^2=Vo^2+2a ×=v*t 


#13
Dec2212, 07:08 PM

P: 217

i used the equation a^2 = b^2 + c^2 To find the distance from A TO D like this a=√((100m)^2 + (100m)^2) I got a = 141.4213562 m I used this equation to find the v2 since v1 = 0 v2^2 = 2xgxd v2^2 = 2(9.8m/s^2)(141.4213562 m) v2 = 52.6484433 m/s to find the time i used this equation v2 = gxt 52.6484433 m/s / (9.8m/s^2) = t t = 5.372290129 s Now for AC I found the angle from D to C to be 135° Then i used the cosine equation which is a^2 = b^2 + c^2 = 2bcCOSθ The a which which is the length from AC came out to be 223.20191262 m I used this distance in this equation v2^2 = 2gd v2^2 = 2(9.8m/s)(223.2019126 m) V2 = 66.201292262 m/s then i used this equation to find the time v2 = gt v2 = (66.201292262 m/s)(t) t = 5.372290132 s Now for the straight line from D to C i used this equation to find the time t = 2x d/v1 i made v2 = 0 t= (2x 100m)/ ( 66.201292262 m/s) t = 3.021060 s 3.0 21060 s + 5.372290132 s = 8.393350889 s But the change between two times.. one for AD and the other one for AC does not match... the answer :( 


#14
Dec2212, 07:10 PM

P: 217




#15
Dec2212, 07:22 PM

P: 17

AD looks good, but you still need to only use the component of gravity acting in the x direction,
Why AC? Solve the problem in two parts: Find the time to go from A to D Find the time to go from D to C Then add the two times together remember there is no force acting on the Car in the x direction from D to C So acceleration is Zero 


#16
Dec2212, 07:27 PM

Homework
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P: 9,849

a. figure out the y component of accn, and use the vertical height in the usual equations b. as above but using horizontal components c. use energy conservation to determine speed at the bottom, then use the hypotenuse distance and the average speed (of start and finish) to get the time. The reason the question mentions a curved section at all is so that you don't have to treat the transition to the horizontal section as an impact. I.e. KE will be conserved. 


#17
Dec2212, 07:51 PM

P: 217




#18
Dec2212, 07:57 PM

P: 17

Gravity is the only force acting on the car in the problem you stated. Since there is a ramp, the for is distributed, Part in the × direction (g*Cos(theta)) and part in the y direction (g*sin(theta))



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