A little cal help, Velocity, displacement, distance?

Question

The velocity function v(t) (in meters per second) is given for a particle moving along a line.
v(t) = t^2-2t-8 1<= t <=5
Find the displacement d1 traveled by the particle during the time interval given above.

And the displacement is -44/3

Find the total distance d2 traveled by the particle during the time interval given above.
Would it not be just be the |-44/3| ?
Or am I a little confused on what they are asking for?

so would this be
|f(1)-f(4)|+|f(4)-f(5)|

Answer 1

Suppose you travel 10 units to the right then 5 units to the left. Your displacement will be 5 units to the right but the distance traveled will be 15 units.

You need to split up the trajectory into parts where the particle is moving left and parts where it moving right. Figure out the distance traveled in each of these parts then add them.

Edit:
Yup! t = 4 is the only time it changes direction on the given time interval.

Answer 2

Displacement, velocity, acceleration are vector parts, it is they have guidelines. Any of them it is contained in the choice direction with appreciate to the axis of reference would be measured negatively. case in point, if a loose falling merchandise is released from relax, for the period of its action the displacement, the fee and the acceleration are all unfavourable with appreciate to a vertical axis pointing upward. If on the choice we % the downward direction to be constructive, then the displacement, the fee, the acceleration would be constructive.

Answer 3

s (t) = ∫ t ² - 2t - 8 dt (between given limits)
s (t) = [ t ³/3 - t ² - 8 t ] between 1 and 5
s (t) = (125/3 - 25 - 40) - (1/3 - 1 - 8)
s (t) = (125/3 - 65) - (1/3 - 3/3 - 24/3)
s (t) = 125/3 - 65 + 26/3
s(t) = 151/3 - 195/3
s (t) = - 44/3 m
d = | - 44/3 | = 44/3 m

Don`t wish to worry you but we appear to agree!