Dolzina poti pri znanem pojemku in hitrosti

v^2 = 2as

Ne da se mi prevajati, še manj pa izpeljevati...
Zamenjaj cifre, ap dobiš rezultat. 20 metrov si že dobil, izračunaj pri višji hitrosti.
[edit] Aja, pa ustavljaš se, ne vstavljaš. Pri tem g-ju se ti lahko kvečjemu kaj vstavi nekam...

If something travels with the initial velocity V.......
1.38m/s

and then stops over the distance L,.....................
0.15m

then, assuming constant decelleration,

L = V^2/2a...........................................
0.15m=1.38m^2/2 x 6.348
0.15=1.9044/12.696


where V - initial velocity
a - absolute value of decceleration.
V^2 means "V squared"

From this, you can find

a = V^2/2L................................................ ......
a=1.38m^2/0.30m
a=6.348m/s

From the Newton's law,

a = F/M................................................. ...
6.348m/s=28566n/4500kg

where F is the force that acted during the slowdown and M is the mass.

We then have

F = MV^2/2L.............................
4500kg x 1.38m/s^2 /2 x 0.15m
4500 x 1.9044 / 0.30m
8569.8/0.30
F=28566n

Use M in kilograms, V in meters per second and L in meters. You'll get the force in Newtons.

Ziher samo toliko? Metrov je več.
(sploh ne znam enačbe napisati preko tipkovnice! Groza... še v excelu bi imel potemtakem težave)