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Math Help - Surds qn

  1. #1
    Member maybeline9216's Avatar
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    Surds qn

    The speed v ms/s of a particle travelling from A to B, at time t s after leaving A, is given by v=10t-t^2.The particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5ms/s or greater for exactly 4sqrt5 s.

    Paper answer:

    v=5 => t1=5-2sqrt5, t2= 5+2sqrt5
    Therefore, t2-t1=4sqrt5

    I dun understand why is it like dat, can someone explain it to me??
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  2. #2
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    Quote Originally Posted by maybeline9216 View Post
    The speed v ms/s of a particle travelling from A to B, at time t s after leaving A, is given by v=10t-t^2.The particle starts from rest at A and comes to rest at B. Show that the particle has a speed of 5ms/s or greater for exactly 4sqrt5 s.

    Paper answer:

    v=5 => t1=5-2sqrt5, t2= 5+2sqrt5
    Therefore, t2-t1=4sqrt5

    I dun understand why is it like dat, can someone explain it to me??
    The speed is not linear; it is not constant; it is not average.

    v = 10t -t^2
    That is a parabola. A vertical parabola that opens downward because of the negative t^2.
    That is because the particle keeps on accelerating up to the mid-distance between A and B, and then it keeps on decelerating right after that mid-distance. (The acceleration and deceleration are equal and opposite and constant.....that's why a parabola.)

    -------------------
    The vertex of that parabola is the maximum speed, and it happens at the mid-distance between A and B.
    There are many ways to find the vertex. (Completing the square, or byCalculus(dv/dt = 0), or by formula "x = -b/2a"

    You should find it at (10,25) .....or, at 10 seconds, and max. V = 25 m/sec.

    ----------------------
    But for the purpose of this qn, the vertex is not needed.

    What are the t's when v = 5 m/sec?

    v = 10t -t^2
    5 = 10t -t^2
    t^2 -10t +5 = 0
    t = {10 +,-sqrt[10^2 -4(1)(5)]} / 2(1)
    t = {10 +,-4sqrt(5)} /2
    t = 5 +,-2sqrt(5) ...........in seconds.

    So, t1 = 5 -2sqrt(5) sec
    t2 = 5 +2sqrt(5) sec.

    That means, see the parabola, between t1 and t2, inclusive, the v is equal or higher than 5 m/sec.
    How long is that in seconds?
    Why, from t1 to t2, of course.
    And that is 4sqrt(5) seconds.
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  3. #3
    Member maybeline9216's Avatar
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    t = {10 +,-sqrt[10^2 -4(1)(5)]} / 2(1)
    t = {10 +,-4sqrt(5)} /2
    t = 5 +,-2sqrt(5) ...........in seconds.


    U used the so called "monster" formula rite??
    Then y do u use the -b-sqrt(b^2-4ac)/2a
    instead of -b+sqrt(b^2-4ac)/2a??
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  4. #4
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    Quote Originally Posted by maybeline9216 View Post
    t = {10 +,-sqrt[10^2 -4(1)(5)]} / 2(1)
    t = {10 +,-4sqrt(5)} /2
    t = 5 +,-2sqrt(5) ...........in seconds.


    U used the so called "monster" formula rite??
    Then y do u use the -b-sqrt(b^2-4ac)/2a
    instead of -b+sqrt(b^2-4ac)/2a??
    Oh, it is called "monster" formula there?
    It is widely known as the Quadratic Formula.

    I used "+,-" ......."plus or minus" .....as it should be. Because there are always two roots from a quadratic equation.
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