NeoPets Riddles (Lenny Conundrums) and Answers Here

deviles: Yes, that's what I meant =)

The text says to ignore any other factors acting against the ships movement, ie. there is no wind resistance making it slow down once it's at top speed.

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shanmyster wrote:

But doesn't the speed increase by 2 meters per second? So he isn't just going 30 meters per second the whole time...

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Majikle wrote:

shanmyster wrote:

But doesn't the speed increase by 2 meters per second? So he isn't just going 30 meters per second the whole time...

Yeah this solution is SO wrong.

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I am probably totally wrong but this is what I did:
4+16+64+100+144+196+256+324+400+484+576+676+784=3548

then i added in the 30*900 for the next 30 acelerations of 30^ and got
27000+3548=30548

This is probably wrong but it's my contribution. I hope it will jog someone's memory

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This is what I did in Excel to get the answer I got:

Col 1 - Seconds (1-60)
Col 2 - Acceleration
Increases from 0-30 in the first 15 seconds
Stays 30 from 16-45 seconds
0 for 46-60 seconds
Col 3 - Velocity
Previous Velocity + Acceleration = Current Velocity
Col 4 - Distance Travelled
Previous Distance + Velocity = Current Distance

If you guys want the screencaps, I can post them

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This might be wrong, but it's what I'm getting so far.

seconds 1-5: travels 30 meters
seconds 6-10: travels 80 meters
seconds 11-15: travels 130 meters

That's a total of 240 meters traveled in the first 15 seconds.

After that his speed stays at a constant of 30 meters per second for the next 30 seconds for a total of 900 meters traveled.

Those 2 added together is 1140 meters traveled, but I've not yet gotten the deceleration rate yet for those last 15 seconds. If his deceleration rate is the same as his acceleration then it might be easier to figure out the distance he travels in his last 15 seconds.

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Seahawker wrote:

This might be wrong, but it's what I'm getting so far.

seconds 1-5: travels 30 meters
seconds 6-10: travels 80 meters
seconds 11-15: travels 130 meters

That's a total of 240 meters traveled in the first 15 seconds.

After that his speed stays at a constant of 30 meters per second for the next 30 seconds for a total of 900 meters traveled.

Those 2 added together is 1140 meters traveled, but I've not yet gotten the deceleration rate yet for those last 15 seconds. If his deceleration rate is the same as his acceleration then it might be easier to figure out the distance he travels in his last 15 seconds.

He doesn't decelerate. In space, a body will remain in motion at the same velocity until something external acts upon it... and we're ignoring external factors

And it's his acceleration that's changing, not speed

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jcrboy wrote:

This is what I did in Excel to get the answer I got:

Col 1 - Seconds (1-60)
Col 2 - Acceleration
Increases from 0-30 in the first 15 seconds
Stays 30 from 16-45 seconds
0 for 46-60 seconds
Col 3 - Velocity
Previous Velocity + Acceleration = Current Velocity
Col 4 - Distance Travelled
Previous Distance + Velocity = Current Distance

If you guys want the screencaps, I can post them

Yes, please =)

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[IMG]This covers the first 48 seconds[/IMG]

I gave you the formulas I used. Hopefully, you can get it from there?

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FORMULA
This is what I did...

Distance traveled during acceleration for 15 seconds

u=starts from rest = 0
a=2m/s2
t=15

First you need to get the maximum speed

v=u+at which is 30m/s

Then get the distance traveled for the 15 seconds acceleration

=(u+v)/2 x t which is 225

Then get the distance traveled at constant acceleration for 30 seconds

constant speed = 30 m/s
t= 30 s

= constant speed x t which is 900

Then get the distance traveled at decelerated to 0 acceleration from constant speed

u= 30 m/s
v=0
t= 15s

=(u+v)/2 x t which is 225

Add it all up and the total distance traveled is 1350m

This problem was take straight out of a physics textbook

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You guys are on right track, but many of you are messing up acceleration for speed. When you add the first 2 seconds of 2+6=8 that gives you speed at 2 seconds. So he actually traveled 10 meters in the first 2 seconds.

I think I'm right, but if anyone knows better I'm open to opinions.

Also, when acceleration is constant for the 30 seconds the speed is still increasing at the rate of TIME=15. So its an extra 58 m/s for 15-45.

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Whoops, screwed up the tags. Trying again

http://i123.photobucket.com/albums/o290/JCRBOY/excel1.jpg

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jcrboy wrote:

Seahawker wrote:
This might be wrong, but it's what I'm getting so far.

seconds 1-5: travels 30 meters
seconds 6-10: travels 80 meters
seconds 11-15: travels 130 meters

That's a total of 240 meters traveled in the first 15 seconds.

After that his speed stays at a constant of 30 meters per second for the next 30 seconds for a total of 900 meters traveled.

Those 2 added together is 1140 meters traveled, but I've not yet gotten the deceleration rate yet for those last 15 seconds. If his deceleration rate is the same as his acceleration then it might be easier to figure out the distance he travels in his last 15 seconds.

He doesn't decelerate. In space, a body will remain in motion at the same velocity until something external acts upon it... and we're ignoring external factors

And it's his acceleration that's changing, not speed

If that's the case then he would travel 1590 meters in that minute. The way the question is phrased though indicates his speed changes after the engines are cut off so that's what I'm trying to figure out..

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I'm just confused...

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Seahawker wrote:

jcrboy wrote:
Seahawker wrote:
This might be wrong, but it's what I'm getting so far.

seconds 1-5: travels 30 meters
seconds 6-10: travels 80 meters
seconds 11-15: travels 130 meters

That's a total of 240 meters traveled in the first 15 seconds.

After that his speed stays at a constant of 30 meters per second for the next 30 seconds for a total of 900 meters traveled.

Those 2 added together is 1140 meters traveled, but I've not yet gotten the deceleration rate yet for those last 15 seconds. If his deceleration rate is the same as his acceleration then it might be easier to figure out the distance he travels in his last 15 seconds.

He doesn't decelerate. In space, a body will remain in motion at the same velocity until something external acts upon it... and we're ignoring external factors

And it's his acceleration that's changing, not speed

If that's the case then he would travel 1590 meters in that minute. The way the question is phrased though indicates his speed changes after the engines are cut off so that's what I'm trying to figure out..

If acceleration is zero, then velocity remains constant

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I used calculus and got 38250 m

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markd wrote:

I used calculus and got 38250 m

I too used calculus and didn't. Do you draw any complaints from my excel plot?

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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?

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the only thing is

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

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I have no complaints on the excel formula, except I don't have excel to do it in! Is there a formula that will give me the answer long hand?

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