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Math Help - total dist. traveled

  1. #1
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    total dist. traveled

    An object moves along a line so that its velocity at time  t is  v(t) = 1/2 + sin2t feet per second. Find the displacement and the total distance traveled by the object for  0 <= t <= 3pi/2

    i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by viet View Post
    An object moves along a line so that its velocity at time  t is  v(t) = 1/2 + sin2t feet per second. Find the displacement and the total distance traveled by the object for  0 <= t <= 3pi/2

    i was able to find the displacement which is 3.35619 ft. need help on finding the total distance traveled.
    You already know that, in order to find displacement we integrate the velocity function:
    \Delta x = \int_0^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )

    We do something similar to get the distance. Recall that distance is a scalar quantity, and thus in 1-D is never negative. This leads us to the formula:
    \Delta x = \int_0^{\frac{3\pi}{2}} dt \left | \frac{1}{2} + sin(2t) \right |

    In order to do this integral, we need to find the intervals over which the integrand is negative. (See graph below.) So we need to know when
    \frac{1}{2} + sin(2t) = 0

    sin(2t) = -\frac{1}{2}

    2t = \frac{7 \pi}{6}, \frac{11\pi}{6}

    t = \frac{7 \pi}{12}, \frac{11\pi}{12}

    And we see that the integrand is negative for the interval \left [ \frac{7 \pi}{12}, \frac{11\pi}{12} \right ]

    So:
    \Delta x = \int_0^{\frac{7\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right ) - <br />
\int_{\frac{7 \pi}{12}}^{\frac{11\pi}{12}} dt \left ( \frac{1}{2} + sin(2t) \right ) <br />
 + \int_{\frac{11 \pi}{12}}^{\frac{3\pi}{2}} dt \left ( \frac{1}{2} + sin(2t) \right )

    I get \Delta x = \sqrt{3} + \frac{5\pi}{12} + 1 \approx 4.04105

    -Dan
    Attached Thumbnails Attached Thumbnails total dist. traveled-velocity.jpg  
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  3. #3
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    Hello, viet!

    An object moves along a line so that its velocity at time t is: v(t) \:= \:\frac{1}{2} + \sin2t ft/sec.

    Find the displacement and the total distance traveled by the object for 0 \leq t \leq \frac{3\pi}{2}

    Integrate: . v(t)\:=\:\frac{1}{2} + \sin2t . . . and we have: . s(t)\;=\;\frac{1}{2}t - \frac{1}{2}\cos2t + C

    . . Assume that the initial position is s(0) = 0, then C = 0.

    . . Hence, the position function is: . s(t) \;= \;\frac{1}{2}t - \frac{1}{2}\cos2t

    At t = \frac{3\pi}{2}:\;s\left(\frac{3\pi}{2}\right)\;=\;  \frac{1}{2}\left(\frac{3\pi}{2}\right) - \frac{1}{2}\cos(3\pi) \:=\:\frac{3\pi}{4} + \frac{1}{2} \:\approx\:2.8562


    Therefore, the displacement is \boxed{2.8562} units to the right.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The object could have stopped and reversed direction.
    . . This happens when v(t) = 0.

    Let v(t) \:=\:\frac{1}{2} + \sin2t\;=\;0

    We have: . \sin2t \:=\:-\frac{1}{2}\quad\Rightarrow\quad 2t\:=\:\frac{7\pi}{6},\:\frac{11\pi}{6} \quad\Rightarrow\quad t \:=\:\frac{7\pi}{12},\:\frac{11\pi}{12}<br />

    . . s\left(\frac{7\pi}{12}\right)\:=\:\frac{1}{2}\left  (\frac{7\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{7\pi}{6}\right)\;=\;\fr  ac{7\pi}{24} + \frac{\sqrt{3}}{4} \:\approx\:1.3493

    . . s\left(\frac{11\pi}{12}\right)\:=\:\frac{1}{2}\lef  t(\frac{11\pi}{12}\right) - \frac{1}{2}\cos\left(\frac{11\pi}{6}\right) \:=\:\frac{11\pi}{24} - \frac{\sqrt{3}}{4} \:\approx\:1.0069


    Now we know all about the object's journey.

    \begin{array}{cccc} s(0) & = & 0\\ s(\frac{7\pi}{12}) & = & 1.3493\\ s(\frac{11\pi}{12}) & = & 1.0069\\ s(\frac{3\pi}{2}) & = & 2.8562\end{array}<br />
\begin{array}{ccc} \}\;1.3493\text{ to the right} \\ \}\;0.3424\text{ to the left } \\  \}\;1.8493\text{ to the right}\end{array}

    . . Total distance: . \boxed{3.541\text{ units}}


    But check my work . . . please!
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  4. #4
    Forum Admin topsquark's Avatar
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    There's only one problem with this:
    Quote Originally Posted by Soroban View Post
    . . Assume that the initial position is s(0) = 0, then C = 0.
    We may certainly assume s(0) = 0, but then
    s(0) = \frac{0}{2} - \frac{1}{2}cos(2 \cdot 0) + C = 0

    gives
    C = \frac{1}{2}, not C = 0

    So all your distance calculations are off by 0.5 ft, which brings you back into agreement with my answers. (Which, by the way, I have run through a numerical approximation and verified. I never dare to assume you got your answer wrong! )

    -Dan
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