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anonymous
 one year ago
Find the time intervals for v(t) = t^2  9t + 18 when the particle is
a.going faster
b.slowing down
anonymous
 one year ago
Find the time intervals for v(t) = t^2  9t + 18 when the particle is a.going faster b.slowing down

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so iv go a. (4.5 , infinity) b. (infinity , 4.5) but apparently their is 2 intervals for each

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3see if the graph helps dw:1438632617992:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think you also need to look at the points where the velocity is becoming 0 (x intercepts)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes thats what i was thinking but that wouldn't allow 2 intervals for each

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438633001610:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1You could use acceleration to your advantage just look at where the acceleration is increasing, and then you can use similar way to determine where it's decreasing as the sign will be opposite of the velocity.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1But ganeshies graph is perfect to, as you can tell easily with it xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i used the acceleration graph to find my answer initially, to be honest i still don't know why its wrong

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3From the graph it is clear that the particle's "speed" is decreasing in the interval \((\infty, ~3)\). Then the particle has changed its direction, and its "speed" starts increasing in the interval \((3,~4.5)\). After that, the particle slows down, its "speed" starts decreasing in the interval \((4.5, ~6)\) and becomes \(0\) at \(t=6\). For the rest, the speed keeps increasing.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2speed=velocitydw:1438633412827:dw you could reflect the part that is below the graph about the xaxis

freckles
 one year ago
Best ResponseYou've already chosen the best response.2below the graph I mean below the xaxis

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3That speed graph looks so much clear!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so then it would be a.(3,4.5)U(6,infinity) b.(infinity,3)U(3,6)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2there is one endpoint just a tad off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where are you getting the eight from ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ "okay so then it would be } \\ a.(3,4.5) \cup (6,\infty) \\ b.(\infty,3) \cup (\color{red}{3},6) \\ \text{ " } \\ \text{ I think this 3 is a typeo}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes it is i ment (4.5 , 6)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438633971759:dw