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How to find the time for a curved path ?

by Lolagoeslala
Tags: curved, path, time
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Lolagoeslala
#1
Dec22-12, 08:24 AM
P: 217
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
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lewando
#2
Dec22-12, 08:49 AM
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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.
mfb
#3
Dec22-12, 10:09 AM
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Quote Quote by lewando View Post
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.
I think that is the point of "small" - we can neglect the time the car spends on that curved path, and consider AD and DC only.

Quote Quote by Lolagoeslala
Then i used this equation to find the time from A to D
V2 - V1 = 2gd
to find V2
That looks wrong.

T don't know how to account for the curved path and how should i continue?
Just assume that the car keeps its velocity there.

Sdtootle
#4
Dec22-12, 12:11 PM
P: 17
How to find the time for a curved path ?

Quote Quote by Lolagoeslal

Then i used this equation to find the time from A to D
V2 - V1 = 2gd
to find V2
I agree with ignoring the curved path. In introductory physics, they will usually use the turn small to insinuate it is negligible. As for the above equation, it should read:
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
Lolagoeslala
#5
Dec22-12, 03:51 PM
P: 217
Quote Quote by mfb View Post
I think that is the point of "small" - we can neglect the time the car spends on that curved path, and consider AD and DC only.


That looks wrong.


Just assume that the car keeps its velocity there.
Ok .. so like for the part from D to C .... you would use the acceleration right instead of G or like gravity? .. and add the times together?
mfb
#6
Dec22-12, 04:01 PM
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P: 11,631
For DC, where do you see an acceleration? I would calculate the velocity with energy conservation (or simply use the value calculated at AD)

and add the times together?
Sure.
Lolagoeslala
#7
Dec22-12, 04:07 PM
P: 217
Quote Quote by mfb View Post
For DC, where do you see an acceleration? I would calculate the velocity with energy conservation (or simply use the value calculated at AD)


Sure.
Well its a straight line that the car is travelling at... what do you mean conservation of energy?
mfb
#8
Dec22-12, 04:15 PM
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P: 11,631
If you don't know about energy conservation, don't use it. You can calculate the velocity with other formulas as well.

Well its a straight line that the car is travelling at...
... and?
Lolagoeslala
#9
Dec22-12, 04:17 PM
P: 217
Quote Quote by mfb View Post
If you don't know about energy conservation, don't use it. You can calculate the velocity with other formulas as well.


... and?
ok so i don't need ... acceleration i can still work with gravity? as in g ... in the motion equations.
Sdtootle
#10
Dec22-12, 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)
Lolagoeslala
#11
Dec22-12, 04:56 PM
P: 217
Quote Quote by Sdtootle View Post
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)
So should i just use the v2 i find.. ?
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?
Sdtootle
#12
Dec22-12, 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
Lolagoeslala
#13
Dec22-12, 07:08 PM
P: 217
Quote Quote by Sdtootle View Post
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)
This is for Path AD

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 :(
Lolagoeslala
#14
Dec22-12, 07:10 PM
P: 217
Quote Quote by Sdtootle View Post
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
i thought u only need G... A from A to C
Sdtootle
#15
Dec22-12, 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
haruspex
#16
Dec22-12, 07:27 PM
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Quote Quote by Lolagoeslala View Post
i thought u only need G... A from A to C
There are several ways to determine the time taken to descend the slope, all producing the same answer:
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.
Lolagoeslala
#17
Dec22-12, 07:51 PM
P: 217
Quote Quote by Sdtootle View Post
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
I do not get why are we gonna use gravity for x i thought it was in an angle.. right... ok so for down would it be 9.8 m/s^2 (down) or (sin 9.8 m/s^2)
Sdtootle
#18
Dec22-12, 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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