Question

I need some assistance with some calculus review problems.

The function \[\ v(t) = t^2 - 8t + 15 \] is the velcotiy of a particle moving along the x - axis, where \[\ t \ge 0 \] and the velocity is measured in m/sec.
(a) Determine when the particle is moving to the right, to the left, and stopped.
(b) Find the particle's displacement for \[\ 0 \le t \le 6 \] Justify answers for (a) and (b).
(c) Write but do not evaluate the integral used to find the total distance traveled by the particle for
\[\ 0 \le t \le 6 \]

Answer 1

I know how to do part c at the least...
\[\text{Total Distance} = \int\limits_{0}^{6} | t^2 - 8t + 15| ~dt\]

Answer 2

okay the others are, its moving forward when velocity is positive and backward when its negative so

Answer 3

displacement u want to find the mean velocity and multiply by time

Answer 4

i think. id have to work it out to be sure

Answer 5

Displacement can be computed by integrating \(v(t)\) over \(0\le t\le6\), since \(\int v(t)~dt=s(t)\) and \(s(t)\) is the displacement.

Answer 6

Oh, i just plugged it in and added the values
3 + 15 = 18
gives the same value as I did when integrating it .

Answer 7

so it's moving to the right when v(t) > 0
left when v(t) < 0
and stopped when v(t) = 0
so
to the right would be: t > 5 , t < 3
to the left would be : 3 < t < 5
and stopped would be when t = 3 , 5

Answer 8

is that correct ?

Answer 9

yes

Answer 10

alrighty, thanks both of you.