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Help me understand this acceleration problem

  1. Feb 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Approaching one of the many sharp horizontal turns in the Monaco Grand Prix, an experienced Formula-1 driver slows down from 135 km/h to 55 km/h while rounding the bend in 10 s. If the driver continues to decelerate at this same rate and the radius of the curve is 15 m, what is the acceleration of the car the moment that its speed reaches 55 km/h?

    2. Relevant equations
    I think we're probably going to need these 2:
    [itex] V_{0x} = a_x t = V_x[/itex]
    [itex]\Delta x = (1/2)(V_x + V_{0x})t [/itex]

    3. The attempt at a solution
    Well before I can start this problem I already have a bunch of question. I would like the members of this board to help me with this only, as to not do the problem for me because I wont learn if that happens.

    So let me see if I understand the question correctly:

    Before reaching the turn, the driver decides to slow down. While traveling with the curve, he is slowing down at the same time from 135 km/h to 55 km/h. and it takes him 10 seconds to slow down to 55 km/h?

    Do I have that right? Or does he slow down to 55 km/h before hitting the curve, and while going through the curve he is at a constant 55 km/h the whole time?

    You're given radius = 15 m, using 15 m * pi you get the distance from where the curve starts to where it ends, 47 m, am I correct here?
     
  2. jcsd
  3. Feb 24, 2017 #2
    Then the question goes on to say "If the car continues to decelerate at this same rate and the radius of the curve is 15 m, what is the acceleration of the car the moment that its speed reaches 55 km/h??"

    So it's decelerating after it has already reached 55 km/h? how does that even make sense?
     
  4. Feb 24, 2017 #3

    haruspex

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    That's how I read it.

    You are not told it is a 180 degree turn. The distance will not matter.

    I do not see the difficulty. It was slowing at a constant rate for 10 seconds (we have to assume) and continues to lose speed at the same rate. The 55km/h is just an instantaneous speed.
    The tricky part is that where the question mentions deceleration and acceleration it means two different things. Its references to deceleration are in relation to speed, whereas the question about its acceleration should be taken to mean acceleration as a vector.
     
  5. Feb 24, 2017 #4
    Hmm I think I'm starting to understand. After I finish the current question I'm on i'll be right back to ask you another one :< lol
     
  6. Feb 26, 2017 #5
    Hey guys please tell me if my method is correct:

    I used formula: [itex] a = \frac {v^2}{r} [/itex]

    where r = radius of the curve, = 15 m
    so final velocity is: 55000 meters/hour, or 15.28 m/s,

    plugging in I get [itex] a = (15.28^2)/(15) = 15.56 m/s^2 [/itex] but my book gives me answer : [itex] 15.8 m/s^2 [/itex]

    is my method correct yall? Was my error in significant figures? I feel like there is more to this problem then just plugging into the equation but I am truly lost here. IF anyone can please help.
     
  7. Feb 26, 2017 #6

    haruspex

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    You have only calculated the centripetal acceleration. You are told that it is still losing speed, so there is also tangential acceleration. The two need to be added, vectorially.
     
  8. Feb 26, 2017 #7
    And do I get tangential acceleration by using formula:

    [itex] V_{0x} + a_x t = V_x [/itex]? If so I did this and got [itex] a_{tan} = -2.222 m/s^2 [/itex]

    then I did [itex]\sqrt {a_{tan}^2 + a_{centri}^2 } [/itex] and my answer was 15.4 m/s^2, still not my books answer :/

    Is this correct though?
     
  9. Feb 26, 2017 #8

    haruspex

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    I do not see how adding the tangential component can have reduced the magnitude of the acceleration. Did you make an error in the signs?
     
  10. Feb 26, 2017 #9

    Weird, I did the same thing again and got 15.64. I guess I may have pressed a wrong button on my calculator or something.

    Do you think that answer is correct? It's still not 15.8 m/s^2 but it is quite close..
     
  11. Feb 26, 2017 #10

    haruspex

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    I'm getting 15.72. Try keeping more digits in the intermediate results.
     
  12. Feb 26, 2017 #11
    Okay I also got 15.723 now. But the method is still correct though, right?

    The method to get the answer = find centripetal acceleration, then find tangential acceleration, then you can find your magnitude and direction, correct?
     
  13. Feb 26, 2017 #12

    haruspex

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    Right.
     
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