NeoPets Riddles (Lenny Conundrums) and Answers Here

rurebeccalyn wrote:

not to be pushy, and i know you guys dont have to do this but...

does anyone have the equation for the problem that they know it right FOR SURE?

I don't think there is one equation that solves this, you have to use multiple since it involves figuring out the new accelerations, then the speeds, and then distance.

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I have NO idea how to put this into Excel.. I don't know how much to fill in myself, and how to get it to calculate the rest on it's own.

I hope they teach us this next year when I have math, or I'll be dead at finals..

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Manx wrote:

rurebeccalyn wrote:
not to be pushy, and i know you guys dont have to do this but...

does anyone have the equation for the problem that they know it right FOR SURE?

I don't think there is one equation that solves this, you have to use multiple since it involves figuring out the new accelerations, then the speeds, and then distance.

gahh i hate math!!!!

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Quote:

his ship's acceleration increased by 2 m/s2 every second

Unfortunately, their wording seems to indicate a logical conclusion of jerk function

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I assumed continuous thrust and continuous acc. etc.

so the 3rd derivative for the first 15 seconds is 2

that makes the 2nd derivative (acceleration) 2t+ C, but C is zero since we started at rest

likewise, the 1st derivative (speed) is t^2 + C, but again C is zero

then the position function is (1/3)t^3, which gives you 1125 when you plug in 15

then, you reset the clock and use the old second derivative to find the new constant acceleration. 2(15) = 30, so the second derivative for the next 30 seconds is 30

that makes the first derivative 30t +c, and the c is determined by the old first derivative at 15. Since 15^2 = 225, the first derivative is 30t+225

the position function at that point becomes 15t^2+225t+c, where the new c is the position we got from the first 15, 1125.

you use 30 seconds as your new t to get 21375 m for the first 45

that leaves the last 15

for those, the new velocity is based on the first derivative from the second scenario, evaluated at 30

30(30)+225 = 1125 m/s

so the first derivative for the last scenario is 1125

that makes the final position function 1125t+c, where the final c is our previous position, meaning 1125t + 21375

this evaluated at 15, gives you 38250

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markd wrote:

you guys are doing it on the assumption that the jerk function works in bursts and I don't know if its a continuous thrust or a bursting thrust

I assumed continuous, since one would think that thrusters wouldn't apply jerk in bursts.

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Quote:

I assumed continuous, since one would think that thrusters wouldn't apply jerk in bursts.

Then my previous post should have the correct answer

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Urgh, I'm going to bed..

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My formulas are as such:

A2-D2 = 0
A3=A2+1 (drag from A3 to A62)
B3=B2+2 (drag from B3 to B17)
B18-B47=B17
B48-B62=0
C3=C2+B3 (drag from C3 to C62)
D3=D2+C3 (drag from D3 to D62)

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waylonjsmithers wrote:

markd wrote:
you guys are doing it on the assumption that the jerk function works in bursts and I don't know if its a continuous thrust or a bursting thrust

I assumed continuous, since one would think that thrusters wouldn't apply jerk in bursts.

Ugh. My answer's going to be off by a few points if you're right

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Quote:

Ugh. My answer's going to be off by a few points

well it all depends on whether neopets meant a continuous jerk or one that adds exactly 2 m/s^2 at exactly the start of every second...

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markd wrote:

Quote:
Ugh. My answer's going to be off by a few points

well it all depends on whether neopets meant a continuous jerk or one that adds exactly 2 m/s^2 at exactly the start of every second...

Maybe I'll be lucky and they'll accept dual answers? Have they ever done that before?

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I dunno

i just started doing lenny conundrum a few weeks ago

and last week's was the first one I've gotten right

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markd wrote:

I dunno

i just started doing lenny conundrum a few weeks ago

and last week's was the first one I've gotten right

I've got a trophy, but it's been a while since I've played. This is the best resource for LC discussion

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FORMULA
For those who want the formulas to this is a a link to a google books page of a textbook that almost has the same problem

http://books.google.com/books?id=oCtw-LWBdP4C&pg=PA120&lpg=PA120&dq=distance+traveled+2+%22m+s2%22&source=web&ots=tlI6iAbf-W&sig=9KSY17MJMZwIEC5hO_-aH3Ol9AI&hl=en

Its a resource for those who want to do it on their own. I used the formula from here and got 1350 meters. Best of luck to all.

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Re: FORMULA

rhubhuwee wrote:

For those who want the formulas to this is a a link to a google books page of a textbook that almost has the same problem

<Link>

Its a resource for those who want to do it on their own. I used the formula from here and got 1350 meters. Best of luck to all.

Unfortunately, the referenced page assumes constant acceleration. During the first 15 seconds, we have increasing acceleration, so this doesn't apply. You could use it in the last 45 seconds when the acceleration is constant from 15 to 45 and then again from 45 to 60 when the acceleration is 0 (another constant).

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so???
BTW
It's my bday!!!

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jcrboy wrote:

My formulas are as such:

A2-D2 = 0
A3=A2+1 (drag from A3 to A62)
B3=B2+2 (drag from B3 to B17)
B18-B47=B17
B48-B62=0
C3=C2+B3 (drag from C3 to C62)
D3=D2+C3 (drag from D3 to D62)

I would come up with
C3 = C2+B2+0.5*(B3-B2)

That is, the speed at the end of the next second is:
C2 - the speed at the end of the previous second +
B2 - the speed added by the acceleration at the end of the previous second +
0.5*(B3-B2) - the speed added by the average of the acceleration during that second (since the increase in the acceleration is constant, we can just take half the difference between the previous and the current acceleration values)

Unfortunately, the calculation for D3 would be even worse, since the average velocity over the second is not so easily calculated as the increase in the velocity is not constant.

D18 = D17 + C17 + 0.5*(C18 - C17) will work, since the acceleration is no longer changing, but D3 to D17 is much more complicated.

I think we need to use markd's approach with calculus.

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edstock wrote:

jcrboy wrote:
My formulas are as such:

A2-D2 = 0
A3=A2+1 (drag from A3 to A62)
B3=B2+2 (drag from B3 to B17)
B18-B47=B17
B48-B62=0
C3=C2+B3 (drag from C3 to C62)
D3=D2+C3 (drag from D3 to D62)

I would come up with
C3 = C2+B2+0.5*(B3-B2)

That is, the speed at the end of the next second is:
C2 - the speed at the end of the previous second +
B2 - the speed added by the acceleration at the end of the previous second +
0.5*(B3-B2) - the speed added by the average of the acceleration during that second (since the increase in the acceleration is constant, we can just take half the difference between the previous and the current acceleration values)

Unfortunately, the calculation for D3 would be even worse, since the average velocity over the second is not so easily calculated as the increase in the velocity is not constant.

D18 = D17 + C17 + 0.5*(C18 - C17) will work, since the acceleration is no longer changing, but D3 to D17 is much more complicated.

I think we need to use markd's approach with calculus.

Unfortunately, I was assuming that he was going 2m/s the entire second second, 6m/s the entire third... etc. I was calculating based on jumps... I could, of course, have decreased the intervals to thousandths of a second to get a more accurate answer. Now I feel silly and that I rushed my answer.

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is it possible to just find the average acceleration for the first 15 seconds, then use d = Vi*t + 0.5*a*t² ?

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