Physics

[Anonymous No.2099676]

Just getting into basic physics and I have run into a problem.

v/t = a
v = 30m/s
t = 5s

30m/s over 5s = 6m?
Which is distance.

v * t = d
30m/s * 5s = 150m/s^2
Which would be acceleration. What am I doing wrong?

btw using d for distance, others should be obvious.

[Anonymous No.2099689]

>>2099676
Try multiplying where you divided, and vice versa...

[Anonymous No.2099695]

30m/s over 5s = 6m/s^2

30m/s * 5s = 150m

[Anonymous No.2099698]

d = v times t. not d = v/t

[Anonymous No.2099699]

>>2099689
Yes, yes... I know that solves it but why? Technically m/s divided by s is m.

[Anonymous No.2099705]

>>2099676
>What am I doing wrong?

Basic Algebra

\thread

[Anonymous No.2099707]

>>2099699
No, it isn't.
(a/b)*b = a
Need to review working with fractions?

[Anonymous No.2099709]

>>2099698
I meant in this case, look at the equation, v/t = a
velocity over time is acceleration but the answer I get is distance.

[Anonymous No.2099713]

>>2099699
>Technically m/s divided by s is m

No, no its not. How are you getting that?
How fucking old are you?

[Anonymous No.2099714]

>>2099699
No, it isn't.
<div class="math">\frac{a/b}{c} = \frac{a}{bc}</div>
Need to review working with fractions?

[Anonymous No.2099721]

>>2099707
Just in case >>2099699 still doesn't get it since he displays a rudimentary failure to do basic math...

(a/b)/b = (a/b) * 1/(1/b) = (a/b)*b = a

[Anonymous No.2099724]

>>2099699
GTFO! JUST GTFO!

[Anonymous No.2099729]

>>2099721
>(a/b)/b = (a/b) * 1/(1/b)
idontthinkso.jpg

[Anonymous No.2099736]

<span class="math">\displaystyle \frac{\frac{m}{s}}{s} = \displaystyle\frac{\frac{m}{s}}{\frac{s}{1}} = \displaystyle \frac{m}{s}\cdot\frac{1}{s} = \displaystyle\frac{m}{s*s} = \displaystyle\frac{m}{s^{2}}[/spoiler]

[sage No.2099734]

>>2099721
YOU CAN GTFO OUT TOO!

[Anonymous No.2099741]

>>2099699
\mathrm{m \over s}\div\mathrm{s}=\mathrm{m \over s}\times\mathrm{1 \over s}=\mathrm{m \over s^2}

[EK !D96TK2PSsI No.2099746]

>>2099734
>GTFO out
>get the fuck out out
wut?

[Anonymous No.2099762]

>>2099746
<span class="math">Show \ me \ your \ pussy [/spoiler]

[Anonymous No.2099766]

>>2099736
How do you do that/learn to do that?

[Anonymous No.2099767]

>mfw
I realized I was retarded. I've done this. Good god. Just haven't seen it as a fraction OVER something, only with the / sign.

[EK !D96TK2PSsI No.2099784]

>>2099762
there you go.

[Anonymous No.2099795]

>>2099784
<span class="math">{\bf{Fap} \ f_{a}^p \ {\bf{fap}}[/spoiler]

[Anonymous No.2099798]

>>2099784
<span class="math">{\bf{Fap}} \ f_{a}^p \ {\bf{fap}}[/spoiler]

[Anonymous No.2099807]

{hurr}

what am i doing wrong?

[Anonymous No.2099808]

(v-u)/t=a
So if the initial velocity (u) is 0m/s, final velocity (v) is 30m/s and time taken to reach that velocity is 5s, you will end up with:
(30-0)/5=a
a=6m/s^2

To get displacement you would use the equation s=[(v-u)/2]t, where s is displacement, you will end up with:
[(30-0)/2]5=s
s=75m

You should memorize these following formulae:
v=u+at
v^2=u^2+2as
s=vt-(0.5)at^2
s=ut+(0.5)at^2 and last but not least,
s=[(v-u)/2]t


s is displacement
u is initial velocity
v is final velocity
a is acceleration
and t is time.

[Anonymous No.2099821]

somehow I get the feeling I am lucky that I suck in english and don't know what the terms ITT mean

is OP retard?

how much?

[Anonymous No.2099828]

>>2099808
Don't you mean: s= p + [(v-u)/2]t
p = initial position

[Anonymous No.2099846]

>>2099828
Displacement means the distance moved from the original position, so it already factors that in.