How to find the time for a curved path ?

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In summary, the car travels a distance of 9.8 meters per second and takes 52.6484433 milliseconds to complete the journey.
  • #1
Lolagoeslala
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Homework Statement


Al right so i am given this diagram. They are asking about the time for the whole journey.
The car is traveling 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.

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/user/LolaGoesLala/media/Untitled_zps73f3ea8a.jpg.html
 
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  • #2
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
lewando said:
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.

Lolagoeslala said:
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.
 
  • #4
Lolagoeslal said:
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
 
  • #5
mfb said:
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?
 
  • #6
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.
 
  • #7
mfb said:
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 traveling at... what do you mean conservation of energy?
 
  • #8
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 traveling at...
... and?
 
  • #9
mfb said:
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.
 
  • #10
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
Sdtootle said:
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?
 
  • #12
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
Sdtootle said:
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 :(
 
  • #14
Sdtootle said:
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
 
  • #15
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
Lolagoeslala said:
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.
 
  • #17
Sdtootle said:
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 going to 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)
 
  • #18
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))
 
  • #19
Sdtootle said:
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))

So i should use the
y direction (g*sin(theta)) or (g*Cos(theta)) for AD or CD.. well CD would the x component right?
 
  • #20
For AD, I would use cos(theta) (the component in the X direction) as it will give you the velocity you need for segment CDo Since there is no acceleration during CD,there are no sin or cos to worry about

Be sure to convince yourself of this as understanding these concepts will make the rest of your semester considerably easier
 
  • #21
Sdtootle said:
For AD, I would use cos(theta) (the component in the X direction) as it will give you the velocity you need for segment CDo Since there is no acceleration during CD,there are no sin or cos to worry about

Be sure to convince yourself of this as understanding these concepts will make the rest of your semester considerably easier

Well when we did this lesson in class. we were looking at the forces... and because it is going down.. we would divide the gravity into gcosθ and gsinθ, so if we are going to a hill ... then i think we would use the angle... which is between a and c ... which would be 135 deg... and for the gravity component down the slope.. i think it would be (9.8m/s^2)xsin135
 
  • #22
If that's the case then it would be because I misunderstood your sketch as the letters weren't entirely clear to me. Regardless, it sounds like you have the right idea. Good luck.
 
  • #23
Sdtootle said:
If that's the case then it would be because I misunderstood your sketch as the letters weren't entirely clear to me. Regardless, it sounds like you have the right idea. Good luck.

So for the straight line path between C and D ... the velocity would be.. the same as A to C right? /.. and the gravity would be 9.8... so i would make v1 = the constant velocity from A to C ...and find V2 ... and work with that?
 
  • #24
Lolagoeslala said:
So for the straight line path between C and D ... the velocity would be.. the same as A to C right? /.. and the gravity would be 9.8... so i would make v1 = the constant velocity from A to C ...and find V2 ... and work with that?

The speed at which the car travels path CD is constant.
This speed is equal to the final speed at C, after the car has traversed path AC. The speed along AC is not constant. Initially, at A, it is zero. Do you know how to find the speed at C?
 
  • #25
lewando said:
The speed at which the car travels path CD is constant.
This speed is equal to the final speed at C, after the car has traversed path AC. The speed along AC is not constant. Initially, at A, it is zero. Do you know how to find the speed at C?

yeah i would use the equation V2^2- V1^1 = 2ad

So i would find the V2 = speed from A to C
CD
and then i would use this V2^2 - v1^2 = 2ad
V1^1 = the speed found for AC
V2^2 = need to find

right?
 
  • #26
I added comments in blue
Lolagoeslala said:
yeah i would use the equation V2^2- V1^1 = 2ad [V1^2]
So i would find the V2 = speed from A to C [the speed AT C]
CD
and then i would use this V2^2 - v1^2 = 2ad
V1^1 = the speed found for AC [V1^2, speed at A--zero]
V2^2 = need to find the speed at C

right?

So, numerically, what did you get for the speed at C (please show your steps)?
 
Last edited:
  • #27
Please see my edits.
 
  • #28
lewando said:
I added comments in blue


So, numerically, what did you get for the speed at C (please show your steps)?

i found the velocity from AC to be :
v2^2 = 2gd
V2 = 44.27188724 m/s

Velocity from C to D or D to C
V2^2 = 2gd + v1^2
v2^2 = 2(9.8)(100) + (44.27188724 m/s)^2
v2 = 62.60990337 m/s
 
  • #29
Comments in blue
Lolagoeslala said:
i found the velocity from AC to be :
v2^2 = 2gd [not "g", but g*sin(45), as discussed earlier]
V2 = 44.27188724 m/s


Velocity from C to D or D to C
V2^2 = 2gd + v1^2 [no, don't do this. You have already figured out what the speed is for this interval from the last step.]

Re-read post #24.​
 
  • #30
lewando said:
Comments in blue

Re-read post #24.

But here we are looking at a different distance... and its gsinθ
I know u guys said to keep the velocity constant which is this 44.27188724 m/s
...​
 
  • #31
Lolagoeslala said:
But here we are looking at a different distance... and its gsinθ
I know u guys said to keep the velocity constant which is this 44.27188724 m/s
...​


You should avoid doing "..." in your posts. No one knows what that means. Please take the time to complete your thoughts.

Can you please reformulate your last post into one or more questions?​
 
  • #32
lewando said:
You should avoid doing "..." in your posts. No one knows what that means. Please take the time to complete your thoughts.

Can you please reformulate your last post into one or more questions?

Well, i was thinking since between C and D, the distance is 100 m, not the same as from A to C. And the gravity is not 9.8sinθ, but since its a straight line wouldn't it just be 9.8?
 
  • #33
My comments in blue
Lolagoeslala said:
Well, i was thinking since between C and D, the distance is 100 m, not the same as from A to C. [true] And the gravity is not 9.8sinθ, but since its a straight line wouldn't it just be 9.8? [technically speaking, gravity does not change during the course of the car's journey. The acceleration due to gravity affects the motion from A to C, but does not affect the motion from C to D, because of the horizontal nature of the motion]
Are you familiar with free body diagrams? Are you familiar with Newton's first law?
 
  • #34
lewando said:
My comments in blue

Are you familiar with free body diagrams? Are you familiar with Newton's first law?

yes sir i am aware. Oh i see what you are trying to say. If the motion is uniform the gravity does not affect it. But what about the distance difference. Should i just use this equation v2 = at to find the time, not considering the distance change? from CD
 
  • #35
If you are able to find the final velocity at the end of the ramp (I'll call V), then the time from C to D can be found with:
V = CD * t
Where CD is the distance between CD. When on a friction less, horizontal plain; the force due to gravity is equal and opposite to the normal force and no acceleration takes place. Since the plain lacks friction, there is no deceleration. Therefore, velocity stays constant during CD
 
<h2>1. How do I calculate the time for a curved path?</h2><p>To calculate the time for a curved path, you will need to know the distance traveled and the speed of the object. Then, you can use the formula t = d/v, where t is the time, d is the distance, and v is the speed. Keep in mind that this formula assumes a constant speed throughout the curved path.</p><h2>2. Does the shape of the curved path affect the time it takes?</h2><p>Yes, the shape of the curved path can affect the time it takes to travel along it. For example, a sharper curve will require a slower speed to maintain the same centripetal force, resulting in a longer time to travel the same distance compared to a gentler curve.</p><h2>3. How does acceleration impact the time for a curved path?</h2><p>Acceleration can impact the time for a curved path in two ways. First, if the object is accelerating, the time will be longer since it is covering more distance. Second, if the object is accelerating towards the center of the curve, it will travel at a higher speed, resulting in a shorter time compared to a constant speed.</p><h2>4. Can I use the same formula for finding time on a curved path as on a straight path?</h2><p>No, the formula for finding time on a curved path is different from the formula for a straight path. This is because the distance traveled on a curved path is not the same as the distance traveled on a straight path, and the speed may also vary along a curved path.</p><h2>5. Are there any other factors that can affect the time for a curved path?</h2><p>Yes, there are other factors that can affect the time for a curved path. These include the mass and shape of the object, air resistance, and the surface of the path (e.g. a smooth surface will result in a faster time compared to a rough surface). Additionally, external forces such as friction and gravity can also impact the time for a curved path.</p>

1. How do I calculate the time for a curved path?

To calculate the time for a curved path, you will need to know the distance traveled and the speed of the object. Then, you can use the formula t = d/v, where t is the time, d is the distance, and v is the speed. Keep in mind that this formula assumes a constant speed throughout the curved path.

2. Does the shape of the curved path affect the time it takes?

Yes, the shape of the curved path can affect the time it takes to travel along it. For example, a sharper curve will require a slower speed to maintain the same centripetal force, resulting in a longer time to travel the same distance compared to a gentler curve.

3. How does acceleration impact the time for a curved path?

Acceleration can impact the time for a curved path in two ways. First, if the object is accelerating, the time will be longer since it is covering more distance. Second, if the object is accelerating towards the center of the curve, it will travel at a higher speed, resulting in a shorter time compared to a constant speed.

4. Can I use the same formula for finding time on a curved path as on a straight path?

No, the formula for finding time on a curved path is different from the formula for a straight path. This is because the distance traveled on a curved path is not the same as the distance traveled on a straight path, and the speed may also vary along a curved path.

5. Are there any other factors that can affect the time for a curved path?

Yes, there are other factors that can affect the time for a curved path. These include the mass and shape of the object, air resistance, and the surface of the path (e.g. a smooth surface will result in a faster time compared to a rough surface). Additionally, external forces such as friction and gravity can also impact the time for a curved path.

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