Total distance traveled question

  • Thread starter Marioqwe
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Homework Statement



Lets say we have the velocity equation for a particle

v(t) = at^3 - bt^2 + ct^ - d with t between 0 and 5

So, to find its displacement I have to integrate v(t) from 0 to 5, and I understand why.
But if I want to find the total distance traveled, I must find where t is negative and then i integrate according to that. So, I would have something like an integral of v(t) from 0 to 4 minus an integral of v(t) from 4 to 5. But why do we subtract? Can anybody explain that to me? We also consider moving backwards when we calculate the total distance traveled right?

Thanks
 

Answers and Replies

  • #2
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Homework Statement



Lets say we have the velocity equation for a particle

v(t) = at^3 - bt^2 + ct^ - d with t between 0 and 5

So, to find its displacement I have to integrate v(t) from 0 to 5, and I understand why.
But if I want to find the total distance traveled, I must find where t is negative and then i integrate according to that.
No, your interval for t is [0, 5], so t is never negative. You need to find where v(t) is negative, because that's when the particle is moving backwards.

If you integrate v(t) from 0 to 5 you'll get the displacement, which is the distance between the particle's position at time t = 0 and time t = 5. For example, if the particle started at the origin, then moved right 5 units, then back 7 units, its displacement would be 2 units, but the distance travelled would be 12 units.
So, I would have something like an integral of v(t) from 0 to 4 minus an integral of v(t) from 4 to 5. But why do we subtract? Can anybody explain that to me? We also consider moving backwards when we calculate the total distance traveled right?
Yes.
 

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