# Calculate average acceleration without time? [closed]

I have a question about a flea jumping, for which I need to show that the average acceleration is around 1000 $ms^{-2}$. I know that to calculate average acceleration you can use $\frac{dv}{dt}$, however I only have the following information:

At the moment the flea's leg leave the surface its body is raised 0.44 mm and it is moving at a speed of 0.95$ms^{-1}$.

I would appreciate it if somebody could help me show the average acceleration of the flea during take off.

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## closed as off-topic by tpg2114, Qmechanic♦Oct 31 '13 at 21:57

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" – tpg2114, Qmechanic
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Have you tried $v^2 = v_0^2 + 2a(r_0 - r)$ –  mcodesmart Oct 4 '13 at 18:39
@ProgrammingEnthusiast Oh my gosh, nope. I knew it would be something simple like that. Would you like to add that as an answer for me, so I can accept it? –  Andy Oct 4 '13 at 18:43

$v^2 = u^2 + 2as$ for a particle undergoing constant acceleration. In this case pf a varying acceleration, this formula can be used to calculate the "average" acceleration, which represents the total change in velocity over the total change in time.

$v$ represents final velocity - in this case 0.95m/s
$u$ represents initial velocity - in this case 0
$s$ represents displacement - in this case 0.44mm, or in SI units, 0.00044m.

Therefore,

$a = \frac{v^2-u^2}{2s}$
$= \frac{0.95^2}{2\times0.00044}$
$= 1025.57 ms^{-2}$

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Shouldn't $u$ stay at $u=0$? No that it makes any significant difference in the value of $a$... –  User58220 Oct 6 '13 at 0:25
@User58220, yes you're right, I put the wrong number in the formula, i'll edit now. The answer hasn't changed however. –  Mew Oct 6 '13 at 3:30

For a particle moving linearly, in three dimensions in a straight line, with constant acceleration, you can use the following equation

$v^2=v^2_0+2a(r_0−r)$

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+1 Thank you very much for your answer. The only reason I've accepted Chris's answer is that it is more complete –  Andy Oct 5 '13 at 10:08

In case you are wondering where the answer comes from: $$a=\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds}$$ which gives $$ads=vdv$$ integrating gives $$v^2=v_0^2 + 2\int_{s_0}^{s}ads$$ using the fundamental theorem of calculus then $$v^2=v_0^2+2\bar{a}(s-s_0)$$

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