Understanding Flea Acceleration: A Homework Conundrum

In summary: Ok, so the definite integral is just finding the sum of all v(t) over all t?Yes! So in summary, the flea leaves the ground at a speed of .84 m/s and reaches a height of 36 mm after a duration of .0042 seconds.
  • #1
kashiark
210
0

Homework Statement


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?


Homework 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

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?
 
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  • #2
kashiark said:
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?
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?

kashiark said:
If you guys just feel like being overly helpful, perhaps you could explain to me why the area under a curve is the displacement?
Let's get the main question out of the way before we start talking about this.
 
  • #3
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
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
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
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
0=v0²+2(-9.8)(.036)
v0=.84m/s
 
  • #8
kashiark said:
0=v0²+2(-9.8)(.036)
v0=.84m/s
Looks good to me. So, what about part (b)?
 
  • #9
.84=0+200t
t=.0042s
 
  • #10
kashiark said:
.84=0+200t
t=.0042s
Why are you using a = 200 m/s2?
 
  • #11
lol I'm not sure i meant 2*10³ so i guess that would make the answer .00042s
 
  • #12
kashiark said:
lol I'm not sure i meant 2*10³ so i guess that would make the answer .00042s
Looks good to me :approve:

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
You are correct.
 
  • #14
The displacement is given by equation

[tex] dx = v(t) dt[/tex]

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

[tex]\int dx=\int v(t) dt[/tex]

[tex]\Delta x=\int v(t) dt[/tex]

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
Ooooooh, ok I think I get it. Thanks!
 
  • #16
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
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.
 
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FAQ: Understanding Flea Acceleration: A Homework Conundrum

1. What is the Flea Acceleration problem?

The Flea Acceleration problem is a hypothetical scenario that involves a flea jumping off a moving train and trying to reach the back of the train. The question is whether the flea would be able to reach the back of the train or if it would be constantly moving towards the front due to the train's acceleration.

2. Is the Flea Acceleration problem a real scientific problem?

No, the Flea Acceleration problem is not a real scientific problem. It is a theoretical problem that is often used as a thought experiment to help understand concepts related to motion and acceleration.

3. What factors would affect the flea's ability to reach the back of the train?

The flea's ability to reach the back of the train would be affected by the initial speed of the train, the flea's jumping ability, air resistance, and the length of the train. These factors would determine how far the flea could jump and whether it could overcome the train's acceleration.

4. What can we learn from the Flea Acceleration problem?

The Flea Acceleration problem can help us understand the concept of relative motion and how it affects the movement of objects. It also highlights the importance of considering all factors, such as air resistance, when analyzing a problem involving motion.

5. Can the Flea Acceleration problem be applied to real-life situations?

While the Flea Acceleration problem itself is not a real-life situation, the concepts and principles it illustrates can be applied to real-world scenarios. For example, understanding relative motion can be helpful in fields such as engineering, physics, and navigation.

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