Thread: Calculus Motion Problem

If someone could please help me with this problem, I'd greatly appreciate it.

A particle moves along the x-axis so that its velocity at any time t≥0 is given by v(t) = (3t^2) -2t - 1 . The position x(t) is 5 for t = 2.
a) For what values of t, 0 ≤ t ≤ 3, is the particle's instantaneous velocity the same as its average velocity on the closed interval [0,3] ?
b) Find the total distance traveled by the particle from time t = 0 until time
t =3.

Thanks for any help!

Hello, turtle!

A particle moves along the x-axis so that its velocity at any time t ≥ 0
is given by: .v(t) .= .3t² -2t - 1. .The position x(t) is 5 for t = 2.

a) For what values of t, 0 ≤ t ≤ 3, is the particle's instantaneous velocity
the same as its average velocity on the closed interval [0,3]?

b) Find the total distance traveled by the particle from t = 0 to t =3.

(a) The average velocity is the definite integral of v(t) from t = 0 to 3
. . .divided by the width of the interval (3).

. . ∫ (3t² - 2t - 1) dt .= .t³ - t² - t (from 0 to 3) .= .(27 - 9 - 3) - (0 - 0 - 0) .= .15

. . . . Hence, the average velocity is: .15 ÷ 3 .= .5

. . When is v(t) = 5? . 3t² - 2t - 1 .= .5 . → . 3t² - 2t - 6 .= .0
. . . . . . . . . . . . . . . . . . . . . . . . __________
. . . . . . . . . . . . . . . . . . . . 2 ± √2² - 4(3)(-6)
. . Quadratic Formula: .t .= .---------------------
. . . . . . . . . . . . . . . . . . . . . . . . 2(6)
. . . . . . . . . . . . . . . . . .__
. . . . . . . . . . . . . .1 + √19
. . Therefore: .t .= .--------
. . . . . . . . . . . . . . . .3


(b) x(t) .= .∫ (3t² - 2t - 1) dt .= .t³ - t² - t + C

We are told that x(2) = 5: .2³ - 2² - 2 + C .= .5 . → . C = 3

. . Hence: .x(t) .= .t³ - t² - t + 3


The particle stops and changes direction when v(t) = 0.
. . 3t² - 2t - 1 .= .0 . → . t = -1/3, 1

x(0) .= .0³ - 0² - 0 + 3 .= .3 . . . The particle starts at +3.

x(1) .= .1³ - 1² - 1 + 3 .= .2 . . . The particle moves 1 unit to the left.

x(3) .= .3³ - 3² - 3 + 2 .= .18 . . . The particle moves 16 units to the right.


Therefore, total distance is: .1 + 16 .= .17 units