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Kinematics Integration Question


IbTrojan

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I'm studying for my math test and there's this one question I'm just not getting. 

 

A particle moves along a horizontal line with velocity given by v(t) = 2t^2 - 11t +12 where t is greater than or equal to zero. 

 

Find the total distance the particle travels from time 2 seconds to time 5 seconds.

 

The question also asks to find acceleration and "a" and "b" if the particle is moving left for a<t<b. Both of which I have figured out. It's just the total distance I can't figure out. We've done other problems in class and I figured them out. My teacher said just to graph it and write out an expression based on the graph and I tried doing that but...I'm getting a really weird negative number :S 

 

Help would very much be appreciated, thank you!

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I'm studying for my math test and there's this one question I'm just not getting. 

 

A particle moves along a horizontal line with velocity given by v(t) = 2t^2 - 11t +12 where t is greater than or equal to zero. 

 

Find the total distance the particle travels from time 2 seconds to time 5 seconds.

 

The question also asks to find acceleration and "a" and "b" if the particle is moving left for a<t<b. Both of which I have figured out. It's just the total distance I can't figure out. We've done other problems in class and I figured them out. My teacher said just to graph it and write out an expression based on the graph and I tried doing that but...I'm getting a really weird negative number :S

 

Help would very much be appreciated, thank you!

 

Here is the theory about how to solve for the distance (i.e. areas under the velocity-time graph) that you've probably learned from the textbook:

 

post-115475-0-44785500-1424656402_thumb.

 

I don't really know whether this is a calculator or a non-cal question, so i'll solve using both

 

Non-Cal (corresponding to Case 2 Method 1 in the theory):

  • Solve the quadratic equation gif.latex? 2t^2 -11t+12 = 0, we'll get gif.latex? x = \frac{3}{2} or gif.latex? x = 4. And since gif.latex? \frac{3}{2} isn't within the range of 2 --> 5, thus we can ignore that solution
  • Split the range into two parts; the first range is from 2 --> 4, and the 2nd range is from 4 --> 5
  • According to the theory, we basically have to solve the following 2 definite integral to get the total distance travel, which is:

gif.latex? \int_2^4(2t^2 -11t+12)dt + \i

gif.latex?= abs\left( \frac{2}{3}t^3 - \

gif.latex?= \frac{14}{3} +\frac{19}{6} =

 

 

With GDC (corresponding to Case 2 Method 2 in the theory): I'll show the steps using my Ti-84plus

  1. Graph the quadratic function gif.latex? f(t) = \left|2t^2 -11t+12\rig. Pay attention to the absolute value sign
  2. Using the definite-integral command on your GDC to find the area under the graph from 2 to 5
  3. The final result obtained from the GDC is 7.83

post-115475-0-91941700-1424658996.png

 

I hope this gave the right answer. Feel free to ask if there's any further questions!

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Thank you so much!!! I realize now, looking at the way I solve it that the mistake I was making was SUBTRACTING the two integrals rather than adding them. Thanks again Vioh!! :D

 

So it's absolute value because it's the distance traveled and that doesn't need a direction, right?

Edited by IbTrojan
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