/sci/ - Science & Math

[Anonymous No.2602480]

this is driving me insane, for some reason I can't get a good intuition for why this constant acceration equation works:
d= Vi(t) + (at²)/2
I can get it for Vi = 0. And I can even derive it. I know it works, but I just can't see intuitively why... if you shot a projectile up with Vi = 10m/s for example, and a = 10m/s², then after the first second that Vi is going to be useless. yet after 20 seconds Vi(t)= 200. fuck I don't know, i'm thinking too much about this shit but maybe someone can shed a bit of light. ty /sci/

[Anonymous No.2602483]

bumpin'

[Anonymous No.2602497]

>after the first second that Vi is going to be useless.
Sure, after 1 second, the velocity is 0m/s. But if it hadn't been for that initial velocity, it would have been 10m/s downward. What the initial velocity was still matters.

[Anonymous No.2602503]

bitte

[EK !EKFQOBUFnQ No.2602507]

its kinda hard to visualise. but it does work.
we know v = u+at (because somethings speed changes over time depending on the rate of accelleration.)

if we integrate this with respect to time, we get

v dv/dt = ut+ 0.5at^2 + C

and seeing as v dv/dt is the integrated velocity, we know it is the displacement because velocity is displacement/time, and time = t (which is what we integrated with respect to)

therefore we have S = ut + 0.5at^2 + C

and assuming no initial displacement, this constant must be zero.

so it is:

S = ut+0.5at^2 + 0
S = ut+0.5at^2

[Anonymous No.2602510]

>>2602497
I know it still matter but why is it multiplied by time?

[Anonymous No.2602521]

>>2602510
Adding 10m/s to the initial velocity changes the velocity at any later point in time by 10 m/s. And therefore it changes the position by (10 m/s) * t.

[Anonymous No.2602528]

>>2602507
Oh wow. thankyou so much!
I can get a little OCD with maths sometimes and I just have to understand everything...

[Anonymous No.2602530]

Basically Vi(t) is the position where the projectile SHOULD be.

(at²)/2 only tells you the displacement from your inertial frame of reference.

So in a way, (at²)/2 is the useless term.

[Anonymous No.2602537]

>>2602521
Yeah I get that in any other circumstance, but somehow it's hard for me to visualise because after that first second its velocity is in the opposite direction yet that vi*t is still there.. cheers for the help

[Anonymous No.2602538]

>>2602507
>v dv/dt is the integrated velocity,
That's some strange notation you have there.

[Anonymous No.2602544]

just picture it as

t*(Vi+at/2)

consider your bricks shat

[EK !EKFQOBUFnQ No.2602548]

>>2602528
no problem, hun :)

tsch, and some people still say Ek isn't helpful...

[Anonymous No.2602553]

>>2602548
Hey EK, I didn't take credit for your fucking tutorial today.

your welcome.

sincerely, project mayhem

[Anonymous No.2602555]

>>2602548
If you actually contributed this way more often you wouldn't have the reputation of a useless tripfag. That said. noko.

[Anonymous No.2602563]

>>2602544
wow not bad.. ty anon

[EK !EKFQOBUFnQ No.2602566]

>>2602553
you can't troll a troll darling =p

>>2602555
i do contribute like this...somewhat often.
have very mixed reputation...i aint even mad.

[Anonymous No.2602575]

[Anonymous No.2602610]

>>2602507
>Takes credit for basic kinematic problems
>Doesn't understand that it's kid's stuff.
Why won't you take credit for solving 13 times 11?

Honestly, what are you doing on this board?

[Anonymous No.2602621]

>>2602610
It's 143.

[Anonymous No.2602637]

This is actually very easy to visualize.

Imagine that you're on a train which is moving with Vi. That is, you're moving at a velocity which makes the ball to appear stationary in the beginning. Now, in your frame, the ball changes position according to
dist. = ((at)^2)/2

Now, to find out how much the ball moved in the original frame of reference, you multiply Vi with the time t and add it on to dist. .

There you go.