Thursday, November 09, 2006

Test Question #17

So, once you have completed #14, you have the two equations v=-4t^3+12t^2 and a= -12^2+24t. You graph those two equations and find all the zeroes because these points are the points at which the particle is changing from speeding up or slowing down to slowing down or speeding up.

When the particle is slowing down, the velocity will be positive and the acceleration will be negative, or the velocity will be negative when the acceleration is positive. For t<0, velocity is positive and acceleration is negative, so this is part of our final domain. At t=2, acceleration becomes negative while velocity is still positive. But at t=3, velocity becomes negative. So, this can be written 2<t<3. The correct notation for the final domain is t<0 U 2<t<3.


At 5:32 AM, Blogger princessophie said...

I would just like to note that there is also a mathematical way to do this problem:

Find the zeroes for both acceleration and velocity.
then do the number line graph indicating where it is positive and negative, use the top for acceleration and the bottom for velocity. Once you have found where the positives and negatives are for both a and v, compare them. when they are the same (either both positive or both negative) the particle is speeding up. When they have different signs (one is positive and the other negative) the particle is slowing down.

just thought i would put that out there for any of you early morning studiers.


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