Question

suppose the velocity of a particle moving in a line is v(t) =t^2-2t-8. find the displacement and distance traveled from 1<=t<=6. answer is -10/3 and 98/3 respectively, how?

Answer 1

did you try integrating ?

Answer 2

the displacement is the integral of v(t) from t=1 to 6
\[ s= \int_1^6 t^2-2t-8 \ dt\]
because the velocity is both positive and negative over part of the the interval, the particle moves away and moves closer. The integral gives the net result. It could happen that after all the moving, the particle ends up where it started, and the displacement is zero. (not in this case, though)

distance is the total distance moved. If the particle moves +2 and then -2, its displacement will be 0, but its distance will be 4 (it moved a total of 2 units out and 2 units returning, or 4 units of distance)

Answer 3

to find the distance traveled, you should break the integral over intervals where the particle is only moving away and intervals where the particle is moving closer. Integrate these regions separately. then take the absolute value of each result and add them up.

Answer 4

you have
\[ v(t)= t^2-2t-8 = (t-4)(t+2) \]

notice that between t= 1 and t=4, the term (t-4) is negative and (t+2) is positive,
their product is negative (meaning v is negative, and the particle is moving closer)

so integrate from t=1 to t=4, and take the absolute value of the distance found.

from t=4 to t=6 the product is positive.
integrate fromt t=4 to t=6 and add to the previous result to get the total distance traveled.