What is the distance covered by a robotic bug moving along the x-axis?

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Homework Help Overview

The problem involves a robotic bug moving along the x-axis, described by the equation of motion x(t) = 10*t^2 - 4*t - 6. The specific question is about determining the distance covered by the bug between t = 0.1 s and t = 0.4 s.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the difference between distance and displacement, with some suggesting that the bug may have turned around during its motion. There are attempts to calculate distance using integration and derivatives, while others question the necessity of these methods.

Discussion Status

There is ongoing exploration of the concepts of distance versus displacement, with participants providing various insights and alternative ways of thinking about the problem. Some have offered guidance on considering the turning points of the bug's motion, while others express confusion about the implications of the problem's setup.

Contextual Notes

Participants note that the problem does not explicitly state whether the bug turns around, leading to discussions about assumptions that can be made based on the given equation of motion.

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Homework Statement



A robotic bug is moving along the x-axis. Its equation of motion is x(t) = 10*t^2- 4*t - 6

I have a problem with only one question:

What is the distance covered by the bug between t = 0.1 s and t = 0.4 s?

Homework Equations

The Attempt at a Solution


[/B]
The first derivative of 10*t^2- 4*t - 6 is velocity: 20*t-4

So I integrated 20*t-4 and computed the value of the integral from 0.1 to 0.4. I got 0.3 and it is wrong. I have no idea why. :(
 
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Why did you Integrated ??... I mean distance is asked and equation of x is given. So can't we just substitute the value and find the answer.
Can you recheck the equation...
 
I have found the value of displacement without a problem. But this is distance covered.
 
Poetria said:
I have found the value of displacement without a problem. But this is distance covered.
Right, but it's only in one dimension. What has to happen for distance covered and magnitude of displacement to be different?
 
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Poetria said:
The first derivative of 10*t^2- 4*t - 6 is velocity: 20*t-4

So I integrated 20*t-4 and computed the value of the integral from 0.1 to 0.4. I got 0.3 and it is wrong. I have no idea why. :(
You differentiated and then integrated, which really just canceled each other. But hold on to that derivative, it'll be useful :)

The problem appears to be asking about the distance that the bug walked, not the displacement from its original location. That is, the length of the curve from where it started to where it finished. So, how do find the arc length of a curve?
 
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You have already gotten some sound advice. I am just going to offer an alternative way of thinking.

If the displacement is not equal to the distance walked, then the bug must have turned around somewhere. How far did the bug have to walk to get there? How far did the bug have to walk back in order to make the displacement what it is?
 
gneill said:
That is, the length of the curve from where it started to where it finished. So, how do find the arc length of a curve?
Curve?
 
haruspex said:
Curve?
In the general sense. A straight line is just a special case.
 
The bug moves in X axis so it moves in straight line... Next (x) in distance or displacement... So the equation give in equation of distance. And the distance traveled by the bug is equal to displacement as the bug travels in straight line.
 
  • #10
Prapti Bala said:
The bug moves in X axis so it moves in straight line... Next (x) in distance or displacement... So the equation give in equation of distance. And the distance traveled by the bug is equal to displacement as the bug travels in straight line.
If you walk straight across the room and back your displacement is zero (you're back where you started). Yet you have covered some distance while walking...

So distance is not always the same as displacement.
 
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  • #11
If there is no turning back then they are equal, right ...
 
  • #12
Prapti Bala said:
If there is no turning back then they are equal, right ...
When motion is along a straight line that is true. So the question is, does the bug in the problem turn around?
 
  • #13
I don't know... the answer. But if the question doesn't say about turning back so can't be assume that it moves forward ?
 
  • #14
The bug must have turned back then. I got it. I have to digest it.
 
  • #15
What did you get, I am totally confused, Please explain ??
 
  • #16
I think I have figured it out.

I have used the derivative to establish the turning point:

20*t-4<0
t<0.2

I devided the path into two parts:
(10*0.2^2-4*0.2-6)-(10*0.1^2-4*0.1-6) = -0.1

(10*0.4^2-4*0.4-6)-(10*0.2^2-4*0.2-6) = 0.4So the distance covered would be 0.5.
 
  • #17
Poetria said:
So the distance covered would be 0.5

Yes, this looks fine.

Prapti Bala said:
But if the question doesn't say about turning back so can't be assume that it moves forward ?

No, you have been told in the problem exactly how it moves by giving the position as a function of time. You can deduce from this whether it is turning around or not as the OP did.
 
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  • #18
Many thanks. :) :) :)
 
  • #19
Thanks... It did raise my knowledge... Thanks again to all
 

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