anonymous
  • anonymous
For an object whose velocity in ft/sec is given by v(t) = cos(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
\[\int\limits_{1}^{5} \cos(t) dt\]
anonymous
  • anonymous
Solve that ^

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More answers

jim_thompson5910
  • jim_thompson5910
what did you get
ganeshie8
  • ganeshie8
careful, you need to put absolute value for "distance"
anonymous
  • anonymous
yeah
anonymous
  • anonymous
1.8
jim_thompson5910
  • jim_thompson5910
good point, or break up the interval into smaller pieces
dan815
  • dan815
find the area under the curve
anonymous
  • anonymous
So......1.8 it is
anonymous
  • anonymous
And dan please give chris his maple syrup
anonymous
  • anonymous
@Ephemera what a lad
anonymous
  • anonymous
@chris00 no man deserves to live without his syrup.
anonymous
  • anonymous
should be around 2.2..
anonymous
  • anonymous
2.2?
anonymous
  • anonymous
Seriously?
anonymous
  • anonymous
i just graphed it...lel
anonymous
  • anonymous
Let's see what @jim_thompson5910 has to say.
ganeshie8
  • ganeshie8
|dw:1444450101879:dw|
jim_thompson5910
  • jim_thompson5910
\[\Large \int_{1}^{5}|\cos(x)|dx\] \[\Large \left|\int_{1}^{\pi/2}\cos(x)dx\right|+\left|\int_{\pi/2}^{3\pi/2}\cos(x)dx\right|+\left|\int_{3\pi/2}^{5}\cos(x)dx\right|\] \[\Large \left|\color{blue}{\int_{1}^{\pi/2}\cos(x)dx}\right|+\left|\color{red}{\int_{\pi/2}^{3\pi/2}\cos(x)dx}\right|+\left|\color{blue}{\int_{3\pi/2}^{5}\cos(x)dx}\right|\] I'll let you finish
1 Attachment
anonymous
  • anonymous
|dw:1444450118173:dw|
jim_thompson5910
  • jim_thompson5910
the blue areas on the graph can be treated as normal the red area must be forced to be positive
anonymous
  • anonymous
if the question doesn't explicitly say to use algebra, by all means you can use graphical means (says time in an exam) . But always good to use calc techniques.
anonymous
  • anonymous
saves*
anonymous
  • anonymous
2.2 was right

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