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Mathematics 26 Online
OpenStudy (anonymous):

Find the distance traveled by the particle during the given time interval. Please help?? :(

OpenStudy (anonymous):

The velocity function in meters per second for a particle moving along a line is given by

OpenStudy (anonymous):

\[ v(t)=3t^2 -2t +1\]

OpenStudy (anonymous):

0<t<3

OpenStudy (anonymous):

Find the distance traveled by the particle during the given time interval. (it should be 0 is less than equal to t is less than equal to 3)

OpenStudy (anonymous):

Please help!

OpenStudy (anonymous):

You have to integrate the equation with the limits of 0 to 3

OpenStudy (anonymous):

Okay, I thought you integrate it if you want to find the displacement of the particle. But this is asking for distance traveled-isn't that asking for something different? :o

OpenStudy (anonymous):

Since its only in one direction the displacement is the distance traveled

OpenStudy (anonymous):

Ah, okay thanks, I'll try integrating it now

OpenStudy (anonymous):

Let me know if it works for you

OpenStudy (anonymous):

So I got 21 m/s.. ? I'll show my work:

OpenStudy (anonymous):

\[\int\limits 3t^2 -2t +1\] =\[3t^3/3 - 2t + 1\] Then I use fundamental theorem of calculus part II : =\[t^3-t^2+t\] =\[(3^3-3^2+3) - (0^3-0^2+0)\] =(21) - (0) =21 m/s

OpenStudy (anonymous):

okay, that's how I got 21. Is that correct?

OpenStudy (anonymous):

Oops, sorry, I showed the second step wrong. But I integrated 2t and 1 to t^2 and t

OpenStudy (anonymous):

Looks right

OpenStudy (anonymous):

Thats what I got

OpenStudy (anonymous):

Yay. Thanks so much!!

OpenStudy (anonymous):

You bet!

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