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    A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.

    Been stuck on it for atleast 2 hours, please help!
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    (Original post by Raeve)
    A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.

    Been stuck on it for atleast 2 hours, please help!
    2 hours? lol damn...

    Particle 1:
    u=0
    a=3.2

    Particle 2:
    u=3
    a=5.4

    Construct they will overtake eachother when their displacements are equal.

    Two equations in s and t and solve them.
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    (Original post by Raeve)
    A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.

    Been stuck on it for atleast 2 hours, please help!
    Hi. The best place to post A level maths questions is F38 (the Maths Forum),
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    (Original post by RDKGames)
    2 hours? lol damn...

    Particle 1:
    u=0
    a=3.2

    Particle 2:
    u=3
    a=5.4

    Construct they will overtake eachother when their displacements are equal.

    Two equations in s and t and solve them.
    I tried that:
    s = 1/2*3.2*t^2
    s=3t+2.7*t^2

    I then equated them, but never got the correct answer.
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    (Original post by Mr M)
    Hi. The best place to post A level maths questions is F38 (the Maths Forum),
    Oh sorry, it is my first post. I did not realise.
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    (Original post by Raeve)
    Oh sorry, it is my first post. I did not realise.
    No problem. I've asked for the thread to be moved. In the meantime, you are in good hands with RDKGames.
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    (Original post by Mr M)
    No problem. I've asked for the thread to be moved. In the meantime, you are in good hands with RDKGames.
    Oh, good to know, thanks!
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    you have to make two simultaneous equations and solve them. let us start off by imagining the instant the second ball has arrived at O.
    what is the distance of first particle from O? s = (5)^2 * 3.2/2 = 25*1.6
    what is the current speed of particle one? u = at = 5*3.2 = 16
    now since we want to know the time both particles are the same distance from O we equate the equations of distance from point O
    eq of first particle = 25*1.6 + 16t + (3.2t^2)/2
    eq of second particle = 3t + (5.4t^2)/2
    the t you will get by solving the quadratic equation you end up with is actually 5 seconds after the t you want because we started from the instant the second ball was at O to simplify things. so just minus 5 from the t you end up with.
    after you get the t use the first t and input it into either of the two equations and get the distance from O.
    Hope this helped.
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    (Original post by RDKGames)
    2 hours? lol damn...

    Particle 1:
    u=0
    a=3.2

    Particle 2:
    u=3
    a=5.4

    Construct they will overtake eachother when their displacements are equal.

    Two equations in s and t and solve them.
    that is wrong because you assume the second particle is accelerating since t = 0 while that is wrong and you cant form any normal single equation to describe the motion of the second ball as it starts accelerating after a certain time. its best to use my method of considering the two balls at the instant in time that the second particle starts to accelerate and has reached O
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    (Original post by Raeve)
    Oh, good to know, thanks!
    See if you can work out what time the speeds of both particles are equal.
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    (Original post by Student1256)
    you have to make two simultaneous equations and solve them. let us start off by imagining the instant the second ball has arrived at O.
    what is the distance of first particle from O? s = (5)^2 * 3.2/2 = 25*1.6
    what is the current speed of particle one? u = at = 5*3.2 = 16
    now since we want to know the time both particles are the same distance from O we equate the equations of distance from point O
    eq of first particle = 25*1.6 + 16t + (3.2t^2)/2
    eq of second particle = 3t + (5.4t^2)/2
    the t you will get by solving the quadratic equation you end up with is actually 5 seconds after the t you want because we started from the instant the second ball was at O to simplify things. so just minus 5 from the t you end up with.
    after you get the t use the first t and input it into either of the two equations and get the distance from O.
    Hope this helped.
    Oh that is great thanks, so I was pretty much there I just didn't add the 40m onto the equation. Infact why do you add it, because we are using, s=ut+1/2at^2?
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    (Original post by Raeve)
    I tried that:
    s = 1/2*3.2*t^2
    s=3t+2.7*t^2

    I then equated them, but never got the correct answer.
    I've misread the context of the question, sorry.

    It's the fact that the second particle comes in 5 seconds later that's the key you are missing.

    Begin by finding the distance travelled by P1 from the origin within the first 5 seconds, and the speed, then equate the distance travelled by P1 in the first 5 seconds + (the displacement of P1 from 5s onwards) with the displacement of the second particle.

    Then just adapt my method above.
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    (Original post by Raeve)
    Oh that is great thanks, so I was pretty much there I just didn't add the 40m onto the equation. Infact why do you add it, because we are using, s=ut+1/2at^2?
    because that is the distance the particle is already away from O and then we are assuming t = 0 at that instant in time when it's that far from O. if you still dont get it basically s = ut + 0.5at^2 means the distance away from a point when you are at that point at t = 0. but when you are already 40 m away from that point at t = 0 you just add it into the equation.
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    (Original post by Student1256)
    because that is the distance the particle is already away from O and then we are assuming t = 0 at that instant in time when it's that far from O. if you still dont get it basically s = ut + 0.5at^2 means the distance away from a point when you are at that point at t = 0. but when you are already 40 m away from that point at t = 0 you just add it into the equation.
    Oh of course! Thanks a lot. I appreciate the help.
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    (Original post by Student1256)
    because that is the distance the particle is already away from O and then we are assuming t = 0 at that instant in time when it's that far from O. if you still dont get it basically s = ut + 0.5at^2 means the distance away from a point when you are at that point at t = 0. but when you are already 40 m away from that point at t = 0 you just add it into the equation.
    I did that but the result is unsolvable. I assume im doing something wrong. Of course I equated them and got the eqn: 1.1t^2-13t-40=0.
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    (Original post by Raeve)
    Oh of course! Thanks a lot. I appreciate the help.
    just remember the only way to do mechanics is to visualise - thats how i did it at least and i wasn't too bad at it!
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    (Original post by Raeve)
    A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.

    Been stuck on it for atleast 2 hours, please help!
    An alternative plan is to say that the second particle has been travelling for 5 seconds less than the first one, so give them times t and t - 5.
    That gives you s = 1.6t^2 and s = 3(t-5) + 2.7(t-5)^2
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    (Original post by Student1256)
    that is wrong because you assume the second particle is accelerating since t = 0 while that is wrong and you cant form any normal single equation to describe the motion of the second ball as it starts accelerating after a certain time. its best to use my method of considering the two balls at the instant in time that the second particle starts to accelerate and has reached O
    Yeah misread.

    (Original post by Raeve)
    Oh of course! Thanks a lot. I appreciate the help.
    Alternatively, to skip some little working, you could've had:

    s=\frac{1}{2}(3.2)(t+5)^2 which is particle 1 in order to compensate for the late start of particle 2
    s=3t+\frac{1}{2}(5.4)t^2

    and then just solve them.
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    (Original post by Raeve)
    A particle starts from rest at a point O on a straight line and moves along the line with constant acceleration of 3.2m/s^2. Five seconds later a second particle passes through O with velocity 3m/s. and moves along the same line with constant acceleration 5.4m/s^2. Find when the second particle overtakes the first and the distance of the two particles from point O.

    Been stuck on it for atleast 2 hours, please help!
    A slightly different method to what's already been suggested:

    If you say 't' is the time from when the second particle passes through O then at t = 0, the first particle will have been travelling for 5 seconds i.e. t + 5. So you get

    Second particle:

    u= 3
    a = 5.4
    t = t

    First particle:

    u = 0
    a = 3.2
    t = t + 5

    Then you can equate the two displacements.

    EDIT : Well it hadn't been suggested before I started writing this post
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    (Original post by RDKGames)
    Yeah misread.



    Alternatively, to skip some little working, you could've had:

    s=\frac{1}{2}(3.2)(t+5)^2 which is particle 1 in order to compensate for the late start of particle 2
    s=3t+\frac{1}{2}(5.4)t^2

    and then just solve them.
    I tried that, but never got the right answer.
 
 
 
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