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AS Maths OCR - Constant acceleration question Watch

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    A particle travels in a straight line and decelerates uniformly at 2 ms-2. When t = 100 its velocity is -v ms-1 (where u>v>0). The average speed of the particle over the 100 seconds is 62.5 ms-1. Find the values of u and v.
    Answers:
    u = 150
    v = 50

    How do I work this out please
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    (Original post by Mohamed_k200123)
    A particle travels in a straight line and decelerates uniformly at 2 ms-2. When t = 100 its velocity is -v ms-1 (where u>v>0). The average speed of the particle over the 100 seconds is 62.5 ms-1. Find the values of u and v.
    Answers:
    u = 150
    v = 50

    How do I work this out please
    Since you have the average speed, you can work out the distance travelled = avg. speed x distance.

    You now have s.
    You know t.
    And you know a.

    Can you now find a suvat equation you can use?


    Ignore me.
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    (Original post by ghostwalker)
    Since you have the average speed, you can work out the distance travelled = avg. speed x distance.

    You now have s.
    You know t.
    And you know a.

    Can you now find a suvat equation you can use?
    The displacement is different to total distance though if the particle changes direction which it does here. Or am I missing something?

    I got the answer but I feel this problem is deceptively hard unless I'm being stupid which is possible.
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    (Original post by Notnek)
    The displacement is different to total distance though if the particle changes direction which it does here. Or am I missing something?

    I got the answer but I feel this problem is deceptively hard unless I'm being stupid which is possible.
    I assumed average speed, was meant to be average velocity, So, probably oversimplifying it, erroneously. Ignore me.
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    (Original post by Mohamed_k200123)
    A particle travels in a straight line and decelerates uniformly at 2 ms-2. When t = 100 its velocity is -v ms-1 (where u>v>0). The average speed of the particle over the 100 seconds is 62.5 ms-1. Find the values of u and v.
    Answers:
    u = 150
    v = 50

    How do I work this out please
    Have you made any progress with this? Can you please post all your working?

    The first step is to find the total distance from the average speed. You should be able to easily form an equation with u/v using the time and acceleration. You'll then need a second equation since you have two unknowns.
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    (Original post by Mohamed_k200123)
    A particle travels in a straight line and decelerates uniformly at 2 ms-2. When t = 100 its velocity is -v ms-1 (where u>v>0). The average speed of the particle over the 100 seconds is 62.5 ms-1. Find the values of u and v.
    Answers:
    u = 150
    v = 50

    How do I work this out please
    The average speed when you have constant acceleration is simply equal to the initial speed plus the final speed (speed after 100 seconds) divided by 2.
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    This might be a helpful tip:

    The particle goes through a period of instantaneous rest and goes back on itself so you might want to deal with two periods of time (first region: when the particle is released to when it comes to rest, second region: from rest to 100 seconds (AFTER FIRST release), finding the distance travelled by the particle in those seperate time periods.

    Doing that, you can find the total distance (rather than displacement) travelled as you have been given the average speed (note this average speed is the average speed of the two time periods).

    Then as Notnek has stated, you need a second equation as you have two unknowns.

    Spoiler:
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    Second equation is one of the "suvat" equations but be careful as these equations deal with displacement rather than distance.


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    (Original post by TheMindGarage)
    The average speed when you have constant acceleration is simply equal to the initial speed plus the final speed (speed after 100 seconds) divided by 2.
    Then the average speed would be (150+50)/2 = 100 but it's actually 62.5.

    Average speed is total distance over total time. If a particle changes direction then you can't just find the mean of the inital and final speed.
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    Since we've had no further input from the OP, I thought I'd post the method I finally used before I forget.


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    Let the time at which the velocity is 0 be t.

    Distance travelled from time = 0 to t is ....
    Distance travelled from time = t to 100 is ....
    Their sum = avg. speed x 100.

    You have a quadratic in t. Solve.

    Since u > v, t must be > 50.

    Knowing t, then u,v follow easily.

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    (Original post by ghostwalker)
    Since we've had no further input from the OP, I thought I'd post the method I finally used before I forget.


    Spoiler:
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    Let the time at which the velocity is 0 be t.

    Distance travelled from time = 0 to t is ....
    Distance travelled from time = t to 100 is ....
    Their sum = avg. speed x 100.

    You have a quadratic in t. Solve.

    Since u > v, t must be > 50.

    Knowing t, then u,v follow easily.


    That's also the way I did it. If this question was in an M1 exam it would be a bloodbath
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    (Original post by Notnek)
    That's also the way I did it. If this question was in an M1 exam it would be a bloodbath
    Justifiably so.

    I assumed it was an early question on someone's learning path, hence my thinking the average "speed" was a misprint. I think it warrants at least 2 out of 3 stars at the end of the chapter.
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    (Original post by ghostwalker)
    Justifiably so.

    I assumed it was an early question on someone's learning path, hence my thinking the average "speed" was a misprint. I think it warrants at least 2 out of 3 stars at the end of the chapter.
    It's a red question at the end of a chapter which I think means hard

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    (Original post by Notnek)
    Then the average speed would be (150+50)/2 = 100 but it's actually 62.5.

    Average speed is total distance over total time. If a particle changes direction then you can't just find the mean of the inital and final speed.
    Ah yes, forgot about that. You'd have to split the motion into two sections.
 
 
 
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