Distance/displacement integration problem

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The discussion revolves around calculating the displacement and distance traveled by a particle given its velocity function v(t) = t² - 2t - 8 over the interval 1 ≤ t ≤ 6. The integral of the velocity function from 1 to 6 yields a displacement of -10/3 meters, indicating that the particle moves in a negative direction. Participants note that the particle may change direction within the interval, which affects the calculation of distance versus displacement. To find the total distance traveled, one must consider the absolute value of the integral and any changes in direction. Clarification on the definitions of displacement and distance is also sought, emphasizing the need to differentiate between the two concepts.
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Homework Statement




Suppose that v(t)=t2 -2t -8, 1≤t≤6 is the velocity function (in meters per second) of a particle moving along a line. Find a) the displacement and b) the distance traveled by the particle during the given time interval.

Homework Equations





The Attempt at a Solution



so i integrated the velocity function at the interval 1 to 6 and got the answer (-10/3)m. so i have no idea if this is the distance or displacement and if it is one of those, then how would i find the other one?

thanks!
 
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zachem62 said:

Homework Statement




Suppose that v(t)=t2 -2t -8, 1≤t≤6 is the velocity function (in meters per second) of a particle moving along a line. Find a) the displacement and b) the distance traveled by the particle during the given time interval.

Homework Equations





The Attempt at a Solution



so i integrated the velocity function at the interval 1 to 6 and got the answer (-10/3)m. so i have no idea if this is the distance or displacement and if it is one of those, then how would i find the other one?

thanks!

Think about whether the particle can change its direction of travel between times 0 and 6, and if it can, think about what the integral ∫v(t): t=0..6 actually computes.

RGV
 
Ray Vickson said:
Think about whether the particle can change its direction of travel between times 0 and 6, and if it can, think about what the integral ∫v(t): t=0..6 actually computes.

RGV

in the question it says that t can only be between 1 and 6 so i can't use any 0. and it is possible that the particle can change direction and my guess is this means that the integral ∫v(t): t=1 to 6 computes the distance. am i right?

then how would i get displacement? can u give me a hint? i know that its the final position minus initial position but i don't know if I'm supposed to get it by integrating the function or do something else.
 
v(t)=t^2 -2t -8, 1≤t≤6
When t= 6, v(6)= 36- 12- 8= 16.
When t= 1, v(1)= 1- 2- 8= -11.
Now, what was the displacement? (Do you know what "displacement" means?)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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