Question

Hey I am struggling to make my values equal when I solve the same input data using different formulas and its driving me mad. If anyone'e up for having a look at my errors that would be awesome please..??

Answer 1

this user @inkyvoyd will solve your struggles

Answer 2

hey dan815, sounds good.. are you saying he's around or should I message him, sorry im not sure of where he is?

Answer 3

don't know how you get your t^2= ???????? that's is not right

Answer 4

you mean the formula? Yeah I have been worried that that is not correct.
How about finding t time in the other formula; \[S = V _{0}t + \frac{ at ^{2} }{ 2}\]

Answer 5

That formula for, t time, has given the correct value before which is why im confused. It worked when initial velocity was 0. Whereas now that initial velocity is not 0 (its 200m/s) the formula or something is no longer working..??? It still yields the same value as the longer formula that it is derived from tho - HENCE MY CONFUSION!!

Answer 6

Is this definitely wrong???\[t =\sqrt{\frac{ 2S }{ 2V _{0}+a }}\]

Answer 7

would you like help with this question?

Answer 8

that would be really good please?

Answer 9

so it looks like you are throwing a projectile straight up in the air

Answer 10

yes, it is about a rocket. I have managed to answer the whole question in a way that gives me the correct answer which is great, but when I try and double check the values that i solved in an alternative formula, I can not get the correct answer. This is my problem. What if I happened to use the alternate formula to solve the question and got the incorrect value for t, time etc.

Answer 11

Would you mind please seeing what answer you get for t, time, suing this formula;

Answer 12

\[S= V _{0}t + \frac{ at ^{2} }{ 2 }\]

Answer 13

you have a mistake

Answer 14

time = 200/9.81 = 20.387 ~ 20.4 you did that correctly

the mistake is after

Answer 15

$$\rm{\Large{
2S = t(V_o + at)
\\ next~ step~ is~ wrong


}}$$

Answer 16

this is correct:
$$\rm{\Large{
2S = t(V_o + at)\\
t = \frac {2S}{V_o + at }


}}$$

Answer 17

ok I am looking now, THANK YOU!

Answer 18

where you have said this "2S=t(Vo+at)next step is wrong"

what did you do with the '2' that was with the Vo? See below..

This is what I thought was correct for that step;
2S=t(2Vo+at)

Answer 19

apart from that, that's ok. Im trying to isolate the t, to get it by itself so I can solve it for t.
so I multiplied both sides by t, in order to get rid of the t in the base line of the fraction. This is how i ended up with t^2 by itself on the left hand side. Are you saying this is incorrect?

Answer 20

hey

Answer 21

hey...

Answer 22

so you saw the mistake?

Answer 23

no im sorry, im still confused. what is the next step please that you do to get t by itself please?

Answer 24

I multiplied both sides by t, in order to get rid of the t in the base line of the fraction. This is how i ended up with t^2 by itself on the left hand side. Are you saying this is incorrect?

Answer 25

yes im saying that is incorrect

Answer 26

ok. I can not see a way then to get t by itself. Can you help me to do that? Or is that a bad question?

Answer 27

thats a valid question :)

Answer 28

phew :) thanks for coming and helping me - i had a look at a question you were working on just now and I see that this must be so so rudimentary for you, cringe. Thanks for your patience and time, I really do appreciate it

Answer 29

i just want to show you want hapens if you multiply by t


$$
\rm{\Large{
2S = V_oT + at^2
\\
2S = t(V_o + at)\\
t = \frac {2S}{V_o + at }\\
t\cdot t = \frac {2S\cdot t }{V_o + at }

}}

$$

ok this doesnt work because the t is stuck in the denominator. squaring doesnt help either .
lets go back a few steps

Answer 30

yes thank you for feeling my pain haha. yes please :D

Answer 31

Can we take it either from here please so I understand what happened to the '2' let me write up what I mean please..

Answer 32

We can go back a few steps.
$$
\rm{\Large{
2S = V_ot + at^2\\
0=- 2S+ V_ot + at^2\\
at^2 + V_ot -2S = 0 \\
This ~ is~ a ~ quadratic~ equation~ in ~t. \\
Use ~ quadratic~ formula ~ to ~ solve~ for ~ t.


\\
}}

$$

Answer 33

right thank you. I have my work cut out for me thats for sure. I will stop trying to apply my wrong wrong wrong formula when trying to solve linear and angular motion etc equations. Oops! Thank you once again :)

Answer 34

I just tried to post a testimonial for you when I became a fan - dont know if it went thru, I hope so, because you rock and everyone should know that :)

Answer 35

I see, wow. Thanks for going and plugging in my values for me so I can see it actually work. I need to do a lot of work around this, and I will :) Thanks again (I have spent longer than I'd like to admit trying and trying to get that other stuff to work out and it was just never going to the way that I was going about it. Thank you)

Answer 36

We can go back a few steps.
$$
\rm{\Large{
2s = v_ot + at^2\\
0=- 2s+ v_ot + at^2\\
at^2 + v_ot -2s = 0 \\
This ~ is~ a ~ quadratic~ equation~ in ~t. \\
Use ~ quadratic~ formula ~ to ~ solve~ for ~ t.\\
t = \frac{-v_o\pm \sqrt{v_o^2 - 4a(-2s)}}{2a}\\~\\
t = \frac{-200\pm \sqrt{200^2 - 4(-9.81)(-2\cdot 2038.7)}}{2(-9.81)}\\~\\
t = \frac{-200\pm \sqrt{200^2 - 1600}}{2(-9.81)}
\\~\\

t = \frac{-200\pm \sqrt{38400}}{2(-9.81)}
\\~\\


\\
}}

$$

Answer 37

oh i see a mistake

Answer 38

impossible hahaha ;)

Answer 39

AWESOME - You probably wont believe me but I have been wondering where tat got to :) GO!! THANKS !!!!

Answer 40

i see more mistakes

Answer 41

still crunching..

Answer 42

I think its better if we use exact decimals, that way we dont have to worry about rounding or possible rounding error

Answer 43

we know that t = 200 / 9.81
therefore
s = 2 000 000 / 9.81 exactly (please verify)

Answer 44

we know that t = 200 / 9.81
therefore
s = 2 000 000 / 981 exactly (please verify)

Answer 45

ok so far? we are going to use these exact values

Answer 46

i would have thought that considering 200/9.81
20000 / 981
but once again im sure Im exposing myself :0

Answer 47

but if you say so, I'll follow

Answer 48

to be clear
t = 20,000 / 981
s = 2,000,000 / 981

Answer 49

oh yes, sorry yes.

Answer 50

ok good

Answer 51

I changed my mind, lets use:
t = 200/9.81
s = 20,000 / 9.81

verify that is still correct

Answer 52

that way i dont have to delete to the point that it becomes intractable

Answer 53

Im double checking my exact value for S. I thought it would be S=2000/ 9.81

Answer 54

HHHEEYYYYYY !! :D

Answer 55

$$
\rm{\large{
s = v_ot + \frac{at^2}{2}\\
2s = \color{red}2v_ot + at^2\\
0=- 2s+2v_ot + at^2\\
at^2 + 2v_ot -2s = 0 \\
This ~ is~ a ~ quadratic~ equation~ in ~t. \\
Use ~ quadratic~ formula ~ to ~ solve~ for ~ t.\\
t = \frac{-2v_o\pm \sqrt{(2v_o)^2 - 4a(-2s)}}{2a}\\~\\
t = \frac{-2v_o\pm \sqrt{(2v_o)^2 + 8\cdot as}}{2a}\\~\\
\\ \therefore \\
t = \frac{-2\cdot 200\pm \sqrt{(2\cdot 200)^2 + 8(-9.81)( \frac{20,000}{9.81 })}}{2(-9.81)}\\~\\

\\ ~ \\ t = \frac{-2\cdot 200\pm \sqrt{(2\cdot 200)^2 + 8(-\cancel{9.81})( \frac{20,000}{\cancel {9.81} })}}{2(-9.81)}\\~\\


t = \frac{-400\pm \sqrt{400^2 - 8 \cdot 20,000}}{2(-9.81)}
\\~\\
t = \frac{-400\pm \sqrt{160,000 - 160,000}}{2(-9.81)}
\\~\\ t = \frac{-400\pm \sqrt{0}}{-19.62}

\\~\\ t = \frac{-400}{-19.62} \pm\frac{ 0 }{-19.62}
\\~\\ t = \frac{-400}{-19.62} \pm0
\\~\\ t = \frac{400}{19.62}
\\~\\ t = \frac{2\cdot 200}{2\cdot 9.81}
\\~\\ t = \frac{ 200}{ 9.81}
}}

$$

Answer 56

That is nuts, you are nuts !! :D

Answer 57

:D

Answer 58

make sure you agree that
t = 200/9.81 = 20.3874...
s = 20,000 / 9.81 = 2038.74...

Answer 59

Never say never say never. And you seemed pretty calm the whole time haha :)

Answer 60

yes, agreed :)

Answer 61

whew, there is probably a way we could avoid that , though.
but you just wanted to double check

Answer 62

the exact value I got for S looking back is 2038.735983

Answer 63

Oh man. Yes I did - couldn't sleep without knowing... but had no idea it was going to blow out into that. I just wouldn't get to sleep if you hadn't come busted it lol!! You are honour the word legend in full haha. THANK SO SOO MUCH. I cant believe it turned into that. Its quite awesome :D

Answer 64

:P

Answer 65

Are you puffed out? haha

Answer 66

not many people have the patience to wait for that , you deserve a pat on the back too

Answer 67

You're too kind - I love it :)

Answer 68

I just have such a long way to go - which I don't mind as long as I don't get throttled along the way ;)

Answer 69

i like to check results myself for consistency. its a good feeling

but we should do this problem abstractly, see if it simplifies

Answer 70

ps i was taking grabs of many of the equations as they evolved so I can later study your approach /problem solving techniques. I hope thats ok with you!?

Answer 71

It sounds interesting but beyond me - I dont know that way..?

Answer 72

the way it came up in the first place (trying to form that formula) was trying to have a back-up plan for solving for t, pretty much. t is not around as much heh

Answer 73

more of a cross checking tool

Answer 74

basically all we did was

Answer 75

we went in a circle

Answer 76

but made it back again - thats the impressive bit. My circles dont finish up being round - or joining at the end lol

Answer 77

turned left at albekurky

Answer 78

Assume we know \( \Large \rm a, v, v_o \)
$$
\Large \rm{
v = v_0 + at\\~\\
\\t = \color{blue}{\frac{v-v_o }{a}} \\
\\ s = v_ot + \frac{at^2}{2}\\
\\ \Rightarrow substitute\\~\\
\\ s = v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \left( \color{red}{\frac{v-v_o }{a}} \right)^2}{2}\\



}
$$

Answer 79

lets simplify that expression

Answer 80

hang on please, im still trying to see how you got the blue step from the red step above...

Answer 81

we can do that separately (i used a computer)

Answer 82

lets assume thats true, and we can go back to it

Answer 83

where did the 'a' on the numerator side of the fraction disappear to?

Answer 84

Assume we know \( \Large \rm a, v, v_o \)
$$
\Large \rm{
v = v_0 + at\\~\\
\\t = \color{blue}{\frac{v-v_o }{a}} \\
\\ s = v_ot + \frac{at^2}{2}\\
\\ \Rightarrow substitute\\
\\ s = v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \left( \color{red}{\frac{v-v_o }{a}} \right)^2}{2} = \color{blue}{\frac{1}{2} \cdot \frac{v^2-v_o^2}{a} }
\\~~\\ Now ~solve ~for~ t
\\ s = v_ot + \frac{at^2}{2}\\
\\ 2s = 2v_ot + at^2
\\ 0 = at^2 + 2v_o t - 2s \\
\\
t = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4a (-2s)} } {2a}

\\~\\ ~~\\ Now ~plug~ in~ s\\
\\ t = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4a (-2\cdot \color{blue}{\frac{1}{2} \cdot \frac{v^2-v_o^2}{a} })} } {2a}
\\~\\= \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4\cancel a (-\cancel 2\cdot \color{blue}{\frac{1}{\cancel 2} \cdot \frac{v^2-v_o^2}{\cancel a} })} } {2 a}
\\~\\ = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4 (-1\cdot \color{blue}{(v^2-v_o^2) })} } {2a}

\\~\\ = \frac{-2v_o \pm \sqrt{ (2v_o)^2 + 4 ( \color{blue}{v^2-v_o^2})} } {2a}
\\~\\ = \frac{-2v_o \pm \sqrt{ 4v_o^2 + 4v^2-4v_o^2} } {2a}
\\~\\ = \frac{-2v_o \pm \sqrt{ 4v^2} } {2a}
\\~\\ = \frac{-2v_o \pm 2v } {2a}
\\~\\ = \frac{-2v_o \pm 2v } {2a}
\\ ~\\ =\frac{2 ( -v_0 \pm v) }{2a}
\\ ~\\ =\frac{ ( -v_0 \pm v) }{a}
}$$

Answer 85

and you see we basically get \( \Large \rm t = \frac{( v - v_0 )}{a} \)

there are actually two solutions for time, because its a quadratic equation.

Answer 86

we get an extraneous solution, in other words

Answer 87

$$
\Large \rm{
v = v_0 + at\\~\\
\\t = \color{blue}{\frac{v-v_o }{a}} \\
\\ s = v_ot + \frac{at^2}{2}\\
\\ \Rightarrow substitute\\
\\ s = v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \left( \color{red}{\frac{v-v_o }{a}} \right)^2}{2} = \color{blue}{\frac{1}{2} \cdot \frac{v^2-v_o^2}{a} }
\\~~\\ Now ~solve ~for~ t
\\ s = v_ot + \frac{at^2}{2}\\
\\ 2s = 2v_ot + at^2
\\ 0 = at^2 + 2v_o t - 2s \\
\\
t = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4a (-2s)} } {2a}

\\~\\ ~~\\ Now ~plug~ in~ s\\
\\ t = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4a (-2\cdot \color{blue}{\frac{1}{2} \cdot \frac{v^2-v_o^2}{a} })} } {2a}
\\~\\= \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4\cancel a (-\cancel 2\cdot \color{blue}{\frac{1}{\cancel 2} \cdot \frac{v^2-v_o^2}{\cancel a} })} } {2 a}
\\~\\ = \frac{-2v_o \pm \sqrt{ (2v_o)^2 - 4 (-1\cdot \color{blue}{(v^2-v_o^2) })} } {2a}

\\~\\ = \frac{-2v_o \pm \sqrt{ (2v_o)^2 + 4 ( \color{blue}{v^2-v_o^2})} } {2a}
\\~\\ = \frac{-2v_o \pm \sqrt{ 4v_o^2 + 4v^2-4v_o^2} } {2a}
\\~\\ = \frac{-2v_o \pm \sqrt{ 4v^2} } {2a}
\\~\\ = \frac{-2v_o \pm 2v } {2a}
\\ ~\\ =\frac{2 ( -v_0 \pm v) }{2a}
\\ ~\\ =\frac{ ( -v_0 \pm v) }{a}
\\ ~\\ =\frac{ (\pm v -v_0 ) }{a}
\\ \therefore \\
t = \frac{ (v -v_0 ) }{a} ~, ~ t = \frac{ (-v -v_0 ) }{a}

\\the~ latter~ solution ~is~ extraneous
\\ because ~ it ~ does~ not~ satisfy ~ v = v_0 + at
}

$$

Answer 88

sorry to but in
the formula you want is average speed times (time) 200/2(20.4)=2040 as you had

Answer 89

i just wanted to demonstrate that we are 'going in circles'

Answer 90

hey perl, I gotta go get ready fro work - night shift, Im loathed to leave but I gotta do it, Im cutting it fine as it is. Thank you so so so sooooo much for taking this on and being so good with me :) I owe you one - somehow, im not sure cause you're kind of on top of math heh :) I will be studying and contemplating this for the next foreseeable future lol but seriously, THANK YOU :)

Answer 91

Your welcome :)

Answer 92

I will leave a challenge problem for you , a step in the proof.
Show that left side equals right side, hint start from the left side cancel
or expand , avoid expanding if not necessary :
$$ \Large \rm
v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \left( \color{red}{\frac{v-v_o }{a}} \right)^2}{2} = \color{blue}{\frac{1}{2} \cdot \frac{v^2-v_o^2}{a} }


$$

Answer 93

ok cool. Thank you, leave it with me (for a while please) I hope I can manage it. I will definitely give it my best and hopefully pull it off :) Thanks for the hints too !
ok I better run, have an awesome night :) :)

Answer 94

$$ \Large \rm
{v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \left( \color{red}{\frac{v-v_o }{a}} \right)^2}{2}
\\= v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{a \color{red}{\frac{(v-v_o)^2 }{a^2 }}}{2}

\\= v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{ \color{red}{\frac{(v-v_o)^2 }{a }}}{2}

\\= v_o \left( \color{red}{\frac{v-v_o }{a}} \right) + \frac{ \color{red}{(v-v_o)^2 }}{2a}
\\= \frac{2}{2}\cdot \frac{v_o (v-v_o) }{a } + \frac{ \color{red}{(v-v_o)^2 }}{2a}
\\= \frac{2v_o (v-v_o) }{2a } + \frac{ \color{red}{(v-v_o)^2 }}{2a}
\\= \frac{2v_o (v-v_o) }{2a } + \frac{ \color{red}{(v-v_o)^2 }}{2a}
\\= \frac{2v_o (v-v_o) + (v-v_o)^2}{2a } \\~\\ factor ~out~ v-v_o\\~\\
\\= \frac{(v-v_o) [ 2v_o + (v-v_o)]}{2a }
\\~\\= \frac{(v-v_o) [ 2v_o + v-v_o]}{2a }
\\~\\= \frac{(v-v_o) [ v_o + v]}{2a }
\\~\\= \frac{(v-v_o) [ v + v_o]}{2a }
\\~\\ = \frac{v^2 - v_o v + v_o v - v_o^2}{2a}
\\~\\ = \frac{v^2 - v_o^2}{2a}
}
$$

Answer 95

sorry i couldn't resist :)

Answer 96

Haha you'r so funny :D Awesome - thanks ha, thanks you :)