Find the total distance traveled....

[Jaskaran]

The problem, "A particle moves along a line so that its velocity at time t is v(t)=t^2 -t -6 (measured in meters per second).

The question: Find the distance traveled during this time period.

My problem: I realize that line for the velocity will dip below the x-axis, so I'm at a loss of where to integrate and how to do it.

I know that v(t)=t^2 -t -6 = (t-3)(t+2), and the book says that it v(t)<0 on [1,3] and v(t)>0 on [3,4], but how did they figure this out?

Then they go on to integrate from ?(1 to 4) |v(t)|dt = ?(1 to 3)[-v(t)] + ?(3 to 4)v(t)dt

and thus they get 10.17 m.

Where are these numbers coming from?

Many thanks appreciated!

 

[mmm4444bot]

v(t) = t^2 - 6

I know that v(t) = t^2 - t - 6

Huh ?

 

[Jaskaran]

fixed

 

[soroban]

Hello, Jaskaran!

\(\displaystyle \text{A particle moves along a line so that its velocity at time }t\text{ is: }\:v(t)\:=\:t^2 -t -6\:\text{ (meters/second).}\)

Find the distance traveled during this time period.

I assume it is from \(\displaystyle t = 1\) to \(\displaystyle t = 4\).

My problem: I realize that graph for the velocity will dip below the x-axis, . This is true.
so I'm at a loss of where to integrate and how to do it.

This means that the velocity is sometimes negative. (The particle moves to the left.)

Why should this make you wonder what to integrate . . . and forget how to integrate?

\(\displaystyle \text{I know that: }\;v(t)\:=\:t^2 -t -6 \:=\: (t-3)(t+2)\)

\(\displaystyle \text{The book says: }\; v(t)<0\,\text{ on }[1,3)\,\text{ and }\,v(t)>0\,\text{ on }(3,4].\)

But how did they figure this out?

\(\displaystyle \text{The graph of }\:v \:=\t-3)(t+2)\:\text{ is an up-opening parabola}\)
. . \(\displaystyle \text{with }x\text{-intercepts 3 and -2.}\)

              |
      *       |             *
              |
       *      |            *
        *     |           *
    - - - * - | - - - - * - - - - -
         -2  *|      *  3
              | * *

\(\displaystyle \text{We are concerned with the graph from }t = 1\text{ to }t = 4.\)

              |
              |             *
              |             :
              |            *:
              |   1       * :
      - - - - | - + - - * - + - - - -
              |   :  *  3   4
              |   *

\(\displaystyle \text{We see that }v(t)\text{ is negative on }[1,3)\,\text{ and positive on }(3,4].\)


\(\displaystyle \text{A baby-talk explanation:}\)

\(\displaystyle \text{From }t = 1\text{ to }t = 3\text{, the particle is moving to the left.}\)

\(\displaystyle \text{At }t = 3\text{, the velocity is 0. \;The particle stops.}\)

\(\displaystyle \text{From }t = 3\text{ to }t = 4\text{, the particle is moving to the right.}\)

Then they go on to integrate from ?(1 to 4) |v(t)|dt = ?(1 to 3)[-v(t)] + ?(3 to 4)v(t)dt
and thus they get 10.17 m.

Where are these numbers coming from?

\(\displaystyle \text{To find the position function }s(t)\text{, we integrate the velocity function }v(t).\)

. . \(\displaystyle s(t) \;=\;\int(t^2-t-6)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 - 6t + C\)


\(\displaystyle \text{When }t = 1\!:\;\;s(1) \;=\;\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) - 6(1) + C \;=\;C - \tfrac{37}{6}\)

\(\displaystyle \text{When }t = 3\!:\;\;s(3) \;=\;\tfrac{1}{3}(3^3) - \tfrac{1}{2}(3^2) - 6(3) + C \;=\;C - \tfrac{27}{2}\)

\(\displaystyle \text{When } t = 4\!:\;\;s(4) \;=\;\tfrac{1}{3}(4^3) - \tfrac{1}{2}(4^2) - 6(4) + C \;=\;C - \tfrac{32}{3}\)



More baby-talk:

\(\displaystyle \text{From }t = 1\text{ to }t = 3\text{, the particle moves from }C - \tfrac{37}{6}\,\text{ to }\,C-\tfrac{27}{2}\)
. . \(\displaystyle \text{a distance of: }\,\bigg|(C-\tfrac{27}{2}) - (C - \tfrac{37}{6})\bigg| \;=\; \frac{22}{3}\text{ meters}\)

\(\displaystyle \text{From }t = 3\text{ to }t = 4\text{, the particle moves from }C-\tfrac{27}{2}\,\text{ to }\,C - \tfrac{32}{3}\)
. . \(\displaystyle \text{a distance of: }\,\bigg|(C - \tfrac{32}{3}) - (C - \tfrac{27}{2})\bigg| \;=\;\frac{17}{6}\text{ meters}\)


\(\displaystyle \text{From }t = 1\text{ to }t = 4\text{, the particle moved: }\:\frac{22}{3} + \frac{17}{6} \:=\:\frac{61}{6} \;=\;10.1666\hdots\text{ meters}\)