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Homework Help: How to find the time for a curved path ?

  1. Dec 22, 2012 #1
    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:
  2. jcsd
  3. Dec 22, 2012 #2


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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.
  4. Dec 22, 2012 #3


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    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.
  5. Dec 22, 2012 #4
    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
  6. Dec 22, 2012 #5
    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?
  7. Dec 22, 2012 #6


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

  8. Dec 22, 2012 #7
    Well its a straight line that the car is travelling at... what do you mean conservation of energy?
  9. Dec 22, 2012 #8


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

    ... and?
  10. Dec 22, 2012 #9
    ok so i don't need ... acceleration i can still work with gravity? as in g ... in the motion equations.
  11. Dec 22, 2012 #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)
  12. Dec 22, 2012 #11
    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?
  13. Dec 22, 2012 #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.
  14. Dec 22, 2012 #13
    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 :(
  15. Dec 22, 2012 #14
    i thought u only need G... A from A to C
  16. Dec 22, 2012 #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
  17. Dec 22, 2012 #16


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    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.
  18. Dec 22, 2012 #17
    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)
  19. Dec 22, 2012 #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))
  20. Dec 22, 2012 #19
    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?
  21. Dec 22, 2012 #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
  22. Dec 23, 2012 #21
    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
  23. Dec 23, 2012 #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.
  24. Dec 24, 2012 #23
    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?
  25. Dec 24, 2012 #24


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    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?
  26. Dec 24, 2012 #25
    yeah i would use the equation V2^2- V1^1 = 2ad

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

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