is this correct? if not, what should i do?
how do i do this
sorry mods meant for off topic
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Smart vlr users come pt4
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CryoZanderDerrekEnjoyer [#2]i will look into it
First question
DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
s stands for distance
part1: 0 - 9.5s
s: 1.5 x 9.5^2
s = 135.375
v: 135.375 = v^2 / 6
v: 812.25 = v^2
v = 28.5
------------------------------------
part2: 9.5s - 11.5s
s: -1.5 x 4 + 28.5 x 2
s = 51
part1 + part2 = 135.375 + 51 = 186.375m
CryoZanderDerrekEnjoyer [#2]i will look into it
Second question
DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
GIVEN:
Sue velocity = 33.0m/s
Sue acceleration = -1.90m/s^2
Sue distance into tunnel = 0m
Van velocity = 5.20m/s
Van acceleration = 0m/s^2
Van distance into tunnel = 175m
LOOKING FOR:
distance and time
EQUATION NEEDED: (For a collision to happen, you need to have both vehicles on the same "coordinates". Thats how they collide. So in following question, we just use the basic kinematic equations and set them equal. The question asks how much time is needed. So we solve for t.
Sue_distance = Van_distance
1/2a x t^2 + v0 x t = (1/2a x t^2 + v0 x t) + 175m
-0.95 x t^2 + 33 x t = (5.2 x t) + 175
-0.95t^2 + 33t = 5.2t + 175
-0.95t^2 + 27.8t - 175 = 0
solve quadratic equation: two solutions, you pick the one closer to 0 because thats the earlier/quicker collision
t = 9.16599
Now for distance just use the same basic kinetic formula
s = 1/2a x t^2 + v0 x t
s: -0.95 x 9.16599^2 + 33 x 9.16599
s: -79.81 + 302.477
s = 222.667m
final solutions:
t = 9.16599
s = 222.667m
additional help:
https://www.mathsisfun.com/quadratic-equation-solver.html
https://www.desmos.com/calculator/ysjxt0jazm
lmk if its correct
CryoZanderDerrekEnjoyer [#5]First question
DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
s stands for distance
part1: 0 - 9.5s
s: 1.5 x 9.5^2
s = 135.375v: 135.375 = v^2 / 6
v: 812.25 = v^2
v = 28.5
------------------------------------
part2: 9.5s - 11.5ss: -1.5 x 4 + 28.5 x 2
s = 51part1 + part2 = 135.375 + 51 = 186.375m
I'll add the steps for the first question under your reply since ... , just in case 😉
CryoZanderDerrekEnjoyer [#8]Second question
DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
GIVEN:
Sue velocity = 33.0m/s
Sue acceleration = -1.90m/s^2
Sue distance into tunnel = 0mVan velocity = 5.20m/s
Van acceleration = 0m/s^2
Van distance into tunnel = 175mLOOKING FOR:
distance and time
EQUATION NEEDED: (For a collision to happen, you need to have both vehicles on the same "coordinates". Thats how they collide. So in following question, we just use the basic kinematic equations and set them equal. The question asks how much time is needed. So we solve for t.
Sue_distance = Van_distance
1/2a x t^2 + v0 x t = (1/2a x t^2 + v0 x t) + 175m
-0.95 x t^2 + 33 x t = (5.2 x t) + 175
-0.95t^2 + 33t = 5.2t + 175
-0.95t^2 + 27.8t - 175 = 0solve quadratic equation: two solutions, you pick the one closer to 0 because thats the earlier/quicker collision
t = 9.16599
Now for distance just use the same basic kinetic formula
s = 1/2a x t^2 + v0 x ts: -0.95 x 9.16599^2 + 33 x 9.16599
s: -79.81 + 302.477s = 222.667m
final solutions:
t = 9.16599
s = 222.667m
additional help:
https://www.mathsisfun.com/quadratic-equation-solver.html
https://www.desmos.com/calculator/ysjxt0jazmlmk if its correct
thanks so much bro
spent last 2 days on these things and my teacher loves to not help
im probably gonna ask for more help like this and thanks in advance
CryoZanderDerrekEnjoyer [#8]Second question
DISTANCE: s = 1/2a x t^2 + v0 x t
VELOCITY: s = v^2 - v0^2 / 2a
GIVEN:
Sue velocity = 33.0m/s
Sue acceleration = -1.90m/s^2
Sue distance into tunnel = 0mVan velocity = 5.20m/s
Van acceleration = 0m/s^2
Van distance into tunnel = 175mLOOKING FOR:
distance and time
EQUATION NEEDED: (For a collision to happen, you need to have both vehicles on the same "coordinates". Thats how they collide. So in following question, we just use the basic kinematic equations and set them equal. The question asks how much time is needed. So we solve for t.
Sue_distance = Van_distance
1/2a x t^2 + v0 x t = (1/2a x t^2 + v0 x t) + 175m
-0.95 x t^2 + 33 x t = (5.2 x t) + 175
-0.95t^2 + 33t = 5.2t + 175
-0.95t^2 + 27.8t - 175 = 0solve quadratic equation: two solutions, you pick the one closer to 0 because thats the earlier/quicker collision
t = 9.16599
Now for distance just use the same basic kinetic formula
s = 1/2a x t^2 + v0 x ts: -0.95 x 9.16599^2 + 33 x 9.16599
s: -79.81 + 302.477s = 222.667m
final solutions:
t = 9.16599
s = 222.667m
additional help:
https://www.mathsisfun.com/quadratic-equation-solver.html
https://www.desmos.com/calculator/ysjxt0jazmlmk if its correct
Same as #11 .... <3
RickyIndian [#12]thanks so much bro
spent last 2 days on these things and my teacher loves to not help
im probably gonna ask for more help like this and thanks in advance
bro.... just use chatgpt
333triplethreat [#15]bro.... just use chatgpt
chatgpt is absolutely horrid for maths/physics
this bitch once said 1^2 = 2
CryoZanderDerrekEnjoyer [#19]chatgpt is absolutely horrid for maths/physics
this bitch once said 1^2 = 2
they arent perfect bro, just ask them to see if they made a mistake and they will spot it. I did that many times and it works
RickyIndian [#12]thanks so much bro
spent last 2 days on these things and my teacher loves to not help
im probably gonna ask for more help like this and thanks in advance
all good bro its a pleasure
shoot me a dm whenever u need something
333triplethreat [#20]they arent perfect bro, just ask them to see if they made a mistake and they will spot it. I did that many times and it works
it works but its not 100% reliable
and if youre not very good at the subject you can get confused quite easily
CryoZanderDerrekEnjoyer [#22]it works but its not 100% reliable
and if youre not very good at the subject you can get confused quite easily
yeah i feel u. prob just best to first look at a yt vid then come back tbh.
RickyIndian [#12]thanks so much bro
spent last 2 days on these things and my teacher loves to not help
im probably gonna ask for more help like this and thanks in advance
spent last 2 days on these things and my teacher loves to not help
cherish any help your teacher gives you, because once you get to college they will not care about you, its all about independence and office hours are basically useless unless you use them correctly, time is money in college.
333triplethreat [#23]yeah i feel u. prob just best to first look at a yt vid then come back tbh.
Khan academy 💯
clocksky888 [#14]Same as #11 .... <3
nice work brotha 🙏 what program do u use for this? i heard someting of LaTeX but i have absolutely no idea about it whatsoever
333triplethreat [#26]ngl i dont really like u but
education bonds us all :)
have a nice night bro ❤️
CryoZanderDerrekEnjoyer [#27]nice work brotha 🙏 what program do u use for this? i heard someting of LaTeX but i have absolutely no idea about it whatsoever
Aye, LATEX for which I used a free of charge Overleaf cloud <-- pretty convenient and handy on the go editor.
Now also one word on the solver too... as technically you have the option to use Wolfram's API through GPT-4 by adding Wolfram as a plug-in <-- but in the paid GPT-4 version. The chat GPT part comes just as an interpreter (and pretty f-ing good one as expected) for the solver to break down the wording into algebraic form ... etc.
Regarding capabilities, both Wolfram's Mathematica and Maple were/are pretty good known calculus solvers - I remember using more Mathematica for calc but on occassions would also use Maple for linear algebra (don't recall why I specifically choose Maple over Mathematica for LinAlg though...) too back in the days before GPT. So, they have been known for their ability to get you going, beside Matlab.
Saying I'm incredibly impressed would be an understatement 🤯, as with the above combo, you could get a clear step by step explanations, and If something isn't clear, you have the option to ask for further clarification <-- this was not possible during my times, but HolyyyF... - amazing how it makes self-study so much easier, though not sure how topic of integration would be dealt, and in what details, with said method.
clocksky888 [#29]Aye, LATEX for which I used a free of charge Overleaf cloud <-- pretty convenient and handy on the go editor.
Now also one word on the solver too... as technically you have the option to use Wolfram's API through GPT-4 by adding Wolfram as a plug-in <-- but in the paid GPT-4 version. The chat GPT part comes just as an interpreter (and pretty f-ing good one as expected) for the solver to break down the wording into algebraic form ... etc.
Regarding capabilities, both Wolfram's Mathematica and Maple were/are pretty good known calculus solvers - I remember using more Mathematica for calc but on occassions would also use Maple for linear algebra (don't recall why I specifically choose Maple over Mathematica for LinAlg though...) too back in the days before GPT. So, they have been known for their ability to get you going, beside Matlab.
Saying I'm incredibly impressed would be an understatement 🤯, as with the above combo, you could get a clear step by step explanations, and If something isn't clear, you have the option to ask for further clarification <-- this was not possible during my times, but HolyyyF... - amazing how it makes self-study so much easier, though not sure how topic of integration would be dealt, and in what details, with said method.
Ight thanks a lot for these useful information bro 🙏🙏🙏 Definetely will help me going
terribletyranny [#30]indian in name and dont know physics = questionable
vlr user when he meets someone who asks for school help:
clocksky888 [#29]Aye, LATEX for which I used a free of charge Overleaf cloud <-- pretty convenient and handy on the go editor.
Now also one word on the solver too... as technically you have the option to use Wolfram's API through GPT-4 by adding Wolfram as a plug-in <-- but in the paid GPT-4 version. The chat GPT part comes just as an interpreter (and pretty f-ing good one as expected) for the solver to break down the wording into algebraic form ... etc.
Regarding capabilities, both Wolfram's Mathematica and Maple were/are pretty good known calculus solvers - I remember using more Mathematica for calc but on occassions would also use Maple for linear algebra (don't recall why I specifically choose Maple over Mathematica for LinAlg though...) too back in the days before GPT. So, they have been known for their ability to get you going, beside Matlab.
Saying I'm incredibly impressed would be an understatement 🤯, as with the above combo, you could get a clear step by step explanations, and If something isn't clear, you have the option to ask for further clarification <-- this was not possible during my times, but HolyyyF... - amazing how it makes self-study so much easier, though not sure how topic of integration would be dealt, and in what details, with said method.
Ive looked into the programs... Jesus christ, 1k+ Dollars for Maple??? Thats absolutely ridiculous.
Guess i will stay with wolfram alpha and khan academy lol...
CryoZanderDerrekEnjoyer [#33]Ive looked into the programs... Jesus christ, 1k+ Dollars for Maple??? Thats absolutely ridiculous.
Guess i will stay with wolfram alpha and khan academy lol...
hehe... F-it ... you'd be more than fine with just Wolfram... you can get Maple Calc app for your phone if you need but for unlimited steps you'd need a paid subscription which depending on your budget may be okeyish ... but essentially, with khan's or/and other stuff it should be more than enough.
One more thing though, If you're into tutorial vids similar to khan, check out mathtutordvd guy <-- he's pretty good in explaining some of the complicated topics, putting them into easy to grasp scenarios etc.
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