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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

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2  
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

3 Answers 3

up vote 1 down vote accepted

$v^2 = u^2 + 2as$ 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  
+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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