Physics equation help
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Physics equation help

[From: ] [author: ] [Date: 12-01-01] [Hit: ]
t = time taken, h = distance traversed,> length of arc traversed = radius x (theta) [s = r θ] Note that (theta) i.e. the angle must be in radians.If angular acceleration is uniform,......
Can anyone list me all the kinematics equations?I will vote for the best answer.Thanks in advance and wish you have a happy new year!!!

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KINEMATICS EQUATIONS
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For Linear Motion
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CASE I : For acceleration = 0 m/s^2

> distance = speed x time [ d = s x t ]
Now rearranging this equation, we have,

> speed = distance / time [ s = d/t ]

> time = distance / speed [ t = d/s ]

CASE II: For uniform acceleration

> v = u + at [ 1st equation of motion ]

> h = ut + 1/2 x a x t^2 [ 2nd Equation of motion ]

> v^2 = u^2 + 2ah [ 3rd Equation of motion ]

where, u = initial velocity, v = final velocity, t = time taken, h = distance traversed, a = acceleration

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For Circular Motion
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> length of arc traversed = radius x (theta) [s = r θ]
Note that (theta) i.e. the angle must be in radians.

> angular speed = (2 x pi x radius) / (time taken) [ ω = 2πR / t ]
= (angular displacement) / (time) [Δθ / Δt]

> tangential speed = angular speed x radius [ v = ω x r ]
= arc of circle traversed / time [ v = ds / dt ]

> radial acceleration = (tangential velocity)^2 / (radius) [ α = v^2 / r ]
= (angular velocity)^2 x radius [ α = ω^2 x r ]
Also called "CENTRIPETAL ACCELERATION"
[Extra Point:Corresponding Force [Centripetal force] = mα = m(v^2 / r)

> tangential acceleration = angular acceleration x radius [ a = α x r ]
= tangential velocity / time [ a = dv / dt ]

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If angular acceleration is uniform,

> θ = [(ω1) x t] + [1/2 x α x t^2] (just like h = ut + 1/2 at^2)

> ω2 = ω1 + αt (just like v = u + at)

> (ω2)^2 - (ω1)^2 = 2αθ (just like v^2 - u^2 = 2aS)

where, θ = angular displacement
ω1 = Initial angular velocity
ω2 = final angular velocity
α = angular acceleration

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The ones you ought to know are:

The SUVAT equations... So

v = u + at
v^2 = u^2 = 2as
s = ut + at^2/2
v = s/t

Then:

A = dv/dt
F = ma
W = Fd
P= Fv = W/t
KE = mv^2/2
GPE = mgh

And then a rearrangement or rather the original version...

Fdt = mdv...

And then conservation of momentum...

m1v1 + m2v2 = v(m1+m2)

I think that that is about it...
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