# Flea Acceleration problem

1. Sep 7, 2009

### kashiark

1. The problem statement, all variables and given/known data
A flea develops an acceleration of 2.0*10³m/s² during takeoff. After takeoff the flea reaches a height of 36mm.
a. How fast does the flea leave the ground?
b. How long does the take-off acceleration last?

2. Relevant equations
d=d0 +(v+v0)t/2
d=d0+v0t+at²/2
v²=v0²+2a(d-d0)
g=-9.8m/s²
the 0's mean initial

3. The attempt at a solution
I'm not sure if I'm supposed to incorporate gravity or not because fleas don't have wings, so the take off must just be the fly jumping, but if you incorporate gravity, the acceleration won't be constant, so none of these formulas will work! However, if I don't incorporate gravity, the flee would have a constant acceleration until 33mm when he just suddenly stops. That wouldn't make any sense... help?

If you guys just feel like being overly helpful, perhaps you could explain to me why the area under a curve is the displacement?

2. Sep 7, 2009

### Hootenanny

Staff Emeritus
As I understand the question, the acceleration of the flea already takes into account the effect of gravity. For example, say you're standing on the ground and then jump up in the air. Somehow we measure your acceleration. This measured value of acceleration is what you are given in the question.

In any case, so long as you a fairly close to the ground, your acceleration is constant. If you throw a ball up in the air, once it has left your hand, it will accelerate at -9.81 m/s2 until it hits the ground. I.e. it's acceleration is constant.

Do you understand why?

3. Sep 7, 2009

### kuruman

You can certainly do part (a). If something reaches a maximum height of 36 mm, how fast was it moving when it left the ground. Note that how it got to that speed is irrelevant for part (a). A certain initial speed buys you a certain maximum height.

Say the speed is v m/s. This becomes the final speed in another calculation where the object starts from rest and accelerates at a given acceleration until it reaches speed v. You have the final speed v, you have the acceleration, can you find the time?

4. Sep 7, 2009

### kashiark

If 2.0*10³m/s² was his constant acceleration, he would never stop; in fact, his velocity would keep increasing and increasing wouldn't it? The ball scenario makes sense because its acceleration is negative. It would just keep accelerating until its initial velocity was overcome, and then it would pick up negative velocity until it hit the ground.

5. Sep 7, 2009

### kuruman

The acceleration of 2*103 m/s2 lasts only as long as the flea has the floor to push against. Once it is airborne, that acceleration is replaced by gravity. That is why you need to solve two separate problems of constant acceleration.

6. Sep 7, 2009

### kashiark

Ohhhhhh! Ok, give me a few minutes, and I'll post my work so you guys can tell me if I'm on the right track.

7. Sep 7, 2009

### kashiark

0=v0²+2(-9.8)(.036)
v0=.84m/s

8. Sep 7, 2009

### Hootenanny

Staff Emeritus
Looks good to me. So, what about part (b)?

9. Sep 7, 2009

### kashiark

.84=0+200t
t=.0042s

10. Sep 7, 2009

### Hootenanny

Staff Emeritus
Why are you using a = 200 m/s2?

11. Sep 7, 2009

### kashiark

lol i'm not sure i meant 2*10³ so i guess that would make the answer .00042s

12. Sep 7, 2009

### Hootenanny

Staff Emeritus
Looks good to me

Now, regarding your other question. I assume you know and understand that the area enclosed by a curve can be found by computing an appropriate definite integral. Yes?

13. Sep 7, 2009

### kashiark

You are correct.

14. Sep 7, 2009

### kuruman

The displacement is given by equation

$$dx = v(t) dt$$

where v(t) is the velocity as a function of time. Integrate both sides to get

$$\int dx=\int v(t) dt$$

$$\Delta x=\int v(t) dt$$

The last equation says that the displacement Δx is the area under the velocity vs. time curve. Don't think of it as a surface area measured in m2. The area under this particular curve has units of meters/second (height) times seconds (base) resulting in meters.

15. Sep 7, 2009

### kashiark

Ooooooh, ok I think I get it. Thanks!

16. Sep 7, 2009

### kashiark

Let me just make sure I have this right: The derivative of a P(t) function would be a v(t) function, so the antiderivative of a v(t) function would be P(t), and the definite integral would be just position?

17. Sep 7, 2009

### kuruman

The definite integral would be position at the upper time limit of integration t2 minus position at the lower limit of integration t1 which would make it the displacement over the time interval t2 - t1.

Last edited: Sep 7, 2009