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I am having a little bit of trouble with this following problem -

Suppose a particle has initial velocity and is moving with a constant acceleration. After t seconds its velocity is v. What will its displacement be in 2t seconds?

I have tried to solve this on my own with the help of these equations -

  1. v = u + at ( a is acceleration, u is initial velocity)
  2. s = ut + 1/2 at^2 ( s is the displacement)
  3. v^2 = u^2 + 2as
  4. s = (u+v)*t/2

I tried to replace all the acceleration parameter using equation 1(a = (v-u)/t) in all other equations. I've also tried some other ways but they didn't help.

but I couldn't figure out how to do it.

I wish I could have tagged it as high-school physics, but couldn't find such tag.

Any help is appreciated!!!!!

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closed as off-topic by centralcharge, John Rennie, Qmechanic Nov 2 '13 at 10:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – centralcharge, John Rennie, Qmechanic
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Is the particle starting from rest? This problem is unsolveable if it is not. – Jerry Schirmer Apr 18 '11 at 19:18
@Jerry Schirmer: No it's not starting from rest. You will just have to express the displacement in terms of v and t. – MD Sayem Ahmed Apr 18 '11 at 19:24
It might have been clearer if you had said something like "Suppose a particle has initial velocity $u$ and is ..." – Henry Apr 18 '11 at 19:31
The following question may also be useful to you – Qmechanic Apr 18 '11 at 22:13
up vote 2 down vote accepted

You have actually solved the problem yourself already! The method you described, solving for the acceleration and plugging it into the formula for displacement, work just fine.

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You have found $a = \dfrac{v-u}{t}$.

You now need to use $s = ut + \dfrac{at^2}{2}$ except that where $t$ appears, you need to replace it with $2t$, and remember that $(2t)^2 = 4 t^2$. Only then should you replace $a$ with the expression you have already found.

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The key to this is that you have two equations in five unknowns, $$s_1=\frac{at^2}{2}+v_0t, \qquad s_2=2at^2+2v_0t.$$ You can solve these for any two variables, leaving three still unknown. You definitely want to know $s_2$, so that's one of them. The other variable you want to isolate is $t$, in an equation that doesn't have $s_2$ in it.

So, solve the first equation for $t$, which is quadratic, choose the positive solution, on the assumptions that $v_0>0$, $t>0$, and $a>0$, which gives $t=\frac{-v_0+\sqrt{v_0^2+2as_1}}{a}$, in terms of $v_0$, $a$, and $s_1$. Substitute that into the equation for $s_2$, and do some tidying.

EDIT: You can see in the result that, on the above assumptions, $s_2$ is between $2\!\!\times$ and $4\!\!\times s_1$, depending on whether the initial velocity or the acceleration dominates the calculation. You should also consider the case when the acceleration is in the opposite direction to the direction of the initial velocity.

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What Mark Eichenlaub said. Or, by simply combining your first and second equations with the appropriate values: 2vt (if you're looking for an answer in terms of v and t). Look at that! The u cancelled! That's kind of elegant.

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