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Home > Science > Mathematics   »   Rates of Change Part 2

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Old Aug 22, 2007, 01:41 AM
sim0nz12345
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Rates of Change Part 2
Hi there
I could really use some help on this particular questions

Show that the particle moving with position equation x(t)=24t^2-3t^4 turns arond after 2 seconds ( find values for position at t=0, t=1, t=2 and t=3 seconds and use a 0.01 second interval to show the velocity is 0 at t=2 seconds.)

I know how to find the values of the positions but not the velocity section

Thanks

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Old Aug 22, 2007, 03:02 AM   #2  
Capuchin
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The velocity is the rate of change of position. You know exactly how to do it, it's the same as the other one i showed you.

It wants you to show that at t=2 the sign of velocity changes from positive to negative.

Again, this is a very unintuitive way to lookat it, i think, but you have to live with what you're given
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Old Aug 22, 2007, 03:21 AM   #3  
sim0nz12345
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Hmmmm i see that I got lazy with the questions
Thanks again
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