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Math Help - uniform acceleration question

  1. #1
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    uniform acceleration question

    A particle X travels with a constant velocity of 6ms^{-1} along a straight line. It passes a particle Y which is stationarity. One second later particle Y accelerates at 2ms^{-2} . How long after being passed it take for particle Y to draw level with the particle X? Leave your answer in surd form.

    for particle 'x' u = 6ms^{-1} ,  a = 0ms^{-2}
    t= T seconds.

    and for particle 'y'  u= 0ms^{-1} ,  a = 2ms^{-2} and t= T-1 seconds

    I am pretty sure you use this formula   s=ut + \frac{1}{2} at^{2}

    but I am not sure how to put all this info together and make one equation any help ?

    thanks!
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  2. #2
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    Quote Originally Posted by Tweety View Post
    A particle X travels with a constant velocity of 6ms^{-1} along a straight line. It passes a particle Y which is stationarity. One second later particle Y accelerates at 2ms^{-2} . How long after being passed it take for particle Y to draw level with the particle X? Leave your answer in surd form.

    for particle 'x' u = 6ms^{-1} ,  a = 0ms^{-2}
    t= T seconds.

    and for particle 'y'  u= 0ms^{-1} ,  a = 2ms^{-2} and t= T-1 seconds

    I am pretty sure you use this formula  s=ut + \frac{1}{2} at^{2}

    but I am not sure how to put all this info together and make one equation any help ?

    thanks!
    Take t=0 to refer to the time when the particles are level.

    Then the position of X at time t>0 is:

    x(t)=6t

    and of particle Y at time t>1:

    y(t)=\frac{2 (t-1)^2}{2}=(t-1)^2.

    When they draw level:

    x(t)=y(t)\ \ \ \ \ \ ..(1)

    and t>1.

    Now (1) is a quadratic in t and you want the root which is >1

    .
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  3. #3
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    Quote Originally Posted by Constatine11 View Post
    Take t=0 to refer to the time when the particles are level.

    Then the position of X at time t>0 is:

    x(t)=6t

    and of particle Y at time t>1:

    y(t)=\frac{2 (t-1)^2}{2}=(t-1)^2.

    When they draw level:

    x(t)=y(t)\ \ \ \ \ \ ..(1)

    and t>1.

    Now (1) is a quadratic in t and you want the root which is >1

    .
    I still don't understand how to solve for 't'.
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  4. #4
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    Quote Originally Posted by Tweety View Post
    I still don't understand how to solve for 't'.
    Have you put x(t)=y(t) and expanded? The you have a quadratic in t. Do you know the quadratic formula or the method of completing the square? Then use it.

    .
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