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A question about equations of motion under constant acceleration

  1. Jun 20, 2014 #1
    This is a very basic question.
    We know for a particle moving under constant acceleration we can use both the equations
    [tex] x = \frac{v + v_0}{2}t [/tex] and
    [tex] x = v_0t + \frac{1}{2}at^2 [/tex]If we want to find [itex]t[/itex] form each of these equations, the first one gives only one value but the second one gives two values of [itex]t[/itex].
    So it seems like if the first equation is used one value of [itex]t[/itex] will be lost.

    Now, I can see one of these equation is linear and another one is quadratic; but still, both of those two equations were derived from same principles - how come one turns out to be better than the other?
     
  2. jcsd
  3. Jun 20, 2014 #2

    adjacent

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    No. The two equations are equally good. One of the ##t##'s you get in the second equation should be rejected because it's absurd.

    Getting two solutions does not mean that it's true. Can you ever have a negative t?
     
  4. Jun 20, 2014 #3

    jbriggs444

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    The two equations do not have the same unknowns. The number of solutions consistent with your knowledge of a system will vary depending on what you know about the system.

    In particular, knowing the sign of the final velocity will eliminate one of the two solutions to the second equation.
     
    Last edited: Jun 20, 2014
  5. Jun 20, 2014 #4
    Would you like to reconsider your statement? Try this problem.

    A ball is thrown vertically upwards with a speed 20ms-1 .What is the time elapsed when the ball is at a height 5m ?

    Do you get negative 't' ? Which of the two values of time would you reject ?
     
    Last edited: Jun 20, 2014
  6. Jun 20, 2014 #5
    By the way, I see no matter which equation I use it gives two results for the above problem! And both values are valid.

    Well now I see, if I try to have same unknowns the two equations become same.

    [itex] y = \frac{v_0+v}{2}t [/itex] If I want to eliminate v from here by [itex] v = v_0+at [/itex] then [itex] y = \frac{2v_0 + at}{2}t = v_0t + \frac{a}{2}t^2 [/itex]

    Thank you all.
     
    Last edited: Jun 20, 2014
  7. Jun 20, 2014 #6

    adjacent

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    We can't use the first equation here.
    Using the second equation, I get
    ##t=2-\sqrt{3} \approx 0.28##
    ##t=2+\sqrt{3} \approx 3.72##

    Certainly you will have to reject one t here. The two times I got here are equally correct. It's the time taken when the ball is at 5m above the ground. Obviously, the last 't' is the time taken after the ball got it's peak height and returned to 5m above the ground.
     
  8. Jun 20, 2014 #7

    adjacent

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    You can't use ##s=\frac{1}{2}(u+v)t## here because you don't know the final velocity ##v##. You only know the initial velocity.
     
  9. Jun 20, 2014 #8
    Then why were you suggesting to the OP that one of the values of 't' would be negative and should be rejected ?
     
  10. Jun 20, 2014 #9
    I don't understand why you said so. The first equation can be used by putting v = 20 - 9.8t

    Why?
     
  11. Jun 20, 2014 #10

    Bandersnatch

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    Adjacent's comment is valid only for final positions below the initial position. I.e., throwing a stone from a rooftop and asking when will it hit the ground nets two values for t of which one is negative and needs to be discarded as unphysical.
    It's not true in general, as has been already shown.
     
  12. Jun 20, 2014 #11

    adjacent

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    Yes, you can do it that way but if you do it that way, that equation becomes the second equation. So it's like using the second equation again. The first equation cannot be used directly.


    I thought Tanya was asking about the time taken for the ball to reach 5m,while moving upwards.

    Oh, I see, Bandersnatch. Thanks for clearing that up. Sorry guys. :shy:
    It's been sometime since I last did these kinematics problems
     
  13. Jun 21, 2014 #12

    BruceW

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    hehe, yes exactly. The two equations are the same. So what was your original problem? You seemed to say that [itex] y = v_0t + \frac{a}{2}t^2 [/itex] gave two solutions? OK, so you mean, if you know the value of ##y## and ##v_0## and ##a## then there are two possible values for ##t##? Yes, I agree with this. And for the other form of the equation [itex] y = \frac{v_0+v}{2}t [/itex] you were saying there is only one solution. So you mean if you are told ##v## and ##y## and ##v_0## then there is only one possible value for ##t##? I agree with this too.

    The reason you get two solutions for one case, and only one solution for the other case is because you are using different information each time. To make it clear, for the first equation, you are using ##y## and ##v_0## and ##a## to find ##t## But in the second equation, you are using ##y## and ##v_0## and ##at+v_0## to find ##t##. (since ##v=at+v_0## and you have used ##v##). So it is clear that in the second case, you have used more information, to restrict the possibilities for ##t##.
     
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