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(Kinematics) Choose the correct statements :

  1. Feb 15, 2012 #1
    1. A particle of mass m moves on the x-axis as follows. It starts from rest at t=0 from the point x=0 and comes to rest at t=1 at the point x=1. No other information is available about its motion at intermediate times (0<t<1). If α denotes the instantaneous acceleration of the particle, then
    (A) α cannot remain positive for all t in the interval 0<t<1
    (B) |α| cannot exceed 2 at any point in its path
    (C) |α| must be >4 at some point or points in its path
    (D) α must change sign during the motion but no other assertion can be made with the information given.

    2. Sorry but i think its obvious.

    3. I agree with statement (A) as if α will remain positive all the time, the particle won't stop or in other words there would be no retardation.
    I have other doubt if statement (D) be like this " α must change sign during the motion ", then it would be correct but after adding "but no other assertion can be made with the information given." , it becomes incorrect?
    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    Got to love "pick the true statement" problems... I agree with you in saying that the particle must have a negative acceleration some time during its motion. Now, (a) states that during the motion, it will have a non-positive acceleration. Now this means it may be 0 or negative. While what actually happens is that the particle accelerates(+) and then needs to decelerate(-). This means that a change of sign for acceleration must occur. Another quick note: (a) also lists a time interval of 0[itex]\leq t \leq1[/itex] while the interval given was 0<t<1. I do not know how important this fact is, but it is still a difference.

    Knowing all of this, this will mean that (a) may be false. So I would choose (d). I cannot see any other assertions that can be made given only this data. These are just my thoughts on it. If anyone else would like to weigh in on it to confirm or deny what I have said, that would be nice. I really hate problems where they try to trick you as hard as they can.
  4. Feb 15, 2012 #3
    Sorry but (A) is True and (D) is False. By the way its a IIT-JEE question so you got to think upon it a little deeper.
  5. Feb 15, 2012 #4


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    re: your statement [red above] the interval was defined over 0≤t≤1.
    Firstly they said what was happening at t=0, then they explained what happens at t=1, then referred to 0<t<1 , which covers the whole 0≤t≤1 interval.

    EDIT: The original heading by OP referred to "Choose the correct statements". note that statements is plural. Perhaps that was a typographical error.
  6. Feb 15, 2012 #5
    Yes i agree with you over the open and closed intervals but i have written the same as printed in the book. And yes the thread name is correct. Its a Multiple Choice Multi Correct Question.
  7. Feb 15, 2012 #6


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    Note: part C is also correct. can you see why?

    EDIT: as well as parts (A) and (D) as you originally noted.
  8. Feb 15, 2012 #7
    @ PeterO
    (A) is correct
    (C) is correct too but don't know why? That's why i posted this question.
    (D)- the original one with
    Code (Text):
    but no other assertion can be made with the information given.
    is not correct.
    Would you please explain why (B) is not correct and (C) is?
  9. Feb 15, 2012 #8


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    The least violet way to meet the t=0 and t=1 conditions, is for the body to move that 1 metre during the one second.
    [one other very violent way to get there is to race away to the 100m mark then stop and come back 99m - all in one second!!]

    Drawing a velocity time graph of the trip is the easiest way to see the possibilities for the 1 metre in 1 second scenario required. The area under that graph has to be 1 unit.

    Either close your eyes and imagine the graph - or do a few quick sketches.

    If the body underwent "infinte" acceleration at t = 0 , then infinite de-celeration at t = 1, it would have been possible to spend the rest of the second [basically all of it] travelling at 1 m/s. The condition of the size of acceleration |a| ≥ 4 is certainly met then.

    If the mass is accelerated for a period of time, then immediately changes to de-celeration, the graph is a triangle.
    Area of a triangle is 1/2 * base * height so since the base of the triangle is 1 unit, the height of the triangle has to be 2.

    If the peak speed of 2 occurs at 0.5 seconds, we have an isosceles triangle, with the slope of each side = 4; representing accelerations of 4 and -4 or |a| = 4 both times.

    If the peak speed occurs before 0.5, then the up slope is steeper than above, while the down slope is less steep. That means the initial acceleration is greater than 4.

    If the peak speed occurs later than 0.5, the initial acceleration is less than 4, but the size of the negative acceleration is greater than 4.

    If the velocity time graph is some other shape, in order to lower the top speed, we will have a trapezium, with the initial up-slope and final down slope being steeper than the case with the triangular graph.

    Thus |a| will be =4 or >4 at some stage.

    Option (B) is, in a sense, the opposite of (C). Only one of them can be true.
  10. Feb 16, 2012 #9
    @ PeterO
    Thanks... The imagination is well written...I understood it now...
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