Question

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

Answer 1

understand that velocity is a derivative of a position (displacement) function

Answer 2

So first I took \[\int\limits_{1}^{6}t^2-2t-8dt\]

Answer 3

why?

Answer 4

v(t) = lim ∆t -> 0= [x(t+Δt) - t]÷∆t = dx/dt

Answer 5

I don't know that form. We're supposed to use the Fundamental Theorem of Calculus.

Answer 6

alright, let us assume that your integral set up is correct

integrate your function then apply V(b) - V(a)

Answer 7

for v(b)-v(a)=v(6)-v(1)=-10/3

Answer 8

oh it's minus

Answer 9

\[\frac{ t^3 }{ 3 }- \frac{ 2t^2 }{ 2 } - 8t ] _{1}^{6}\]

Answer 10

I actually think I figured out the problem.

Answer 11

[6^2/3 - 6^2 - 8(6)] - [(1/3-1-8)]

Answer 12

I figured it out. Thank you for helping though! I just needed to take some time to try and work it out.