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V = Vo + at

X - Xo = Vot + .5at2

v2 = vo2 + 2a(X - Xo)

X - Xo = .5(Vo + V)t

V is final velocity in units of meters per seconds (m/s)

Vo is initial velocity in units of meters per seconds (m/s)

a is acceleration in units of meters per squared seconds (m/s2)

t is time in seconds (s)

X is final displacement in units of meters (m)

Xo is initial displacement in units of meters (m)

Multiple unknowns can be solved for each other by using more than one equation. If you are solving for two components, find a common piece (such as time), solve for that piece for each component and set them equal to each other to solve for the other unknowns.

Hope this helps!

p.s. sometimes it is easier to remember the second, third and fourth equations as

2. df - dI = vIt + .5at2

3. v2=vI2+2a(df - di)

4. df - dI = .5(vI + vf)t

its all the same except x has been switched with the first letter of displacement, and the 0 have been switched with i for initial. f is for final

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Q: What are the kinematic equations?

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Simply put, kinematics is really just physics without forces or masses. That is, you deal with velocities, accelerations, time, etc. So a kinematic equation will have those variables.The kinematic equation of motion could be any of the four equations I list, or any variation of them (they can be rewritten in a number of ways):let d = distance, v = velocity, i = initial velocity, a = acceleration, t = timev = i + atd = it + (1/2)t2v2 = i2 + 2add = (1/2)(i + v)tThe equations describe the motion, whether it describing it's acceleration, velocity, distance traveled along a certain axis, all with respect to time.

Open kinematic pair chain exercises are primarily for strengthening and rehabilitating purposes. These exercises allow the hand and/or foot to be moved freely.

The kinematic chain is a combination of four or more kinetic pairs, such that the relative motion between the link is completely constrained. The simplest & basic kinematic chain is a four bar chain.

it should be in the range of 1.3-1.4 at 40 c

Ionic equations differ from chemical equations in that substances that are ions in solution are written as ions in the equation

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The answer is "No". If acceleration changes, forces of inertia should be taken to consideration. It requires dynamic equations of motion. However, if acceleration changes are not significant, you may continue using kinematics. To check if kinematic solution is within required precision limits you need to compare the solution of kinematic and dynamic equations and decide if kinematic solution is good enough.

it not possibl that the eq of kinetic is 1/2 mv2

Acceleration must be constant to use kinematic equations. Acceleration need not be constant if working with energy.

There are four kinematic equations. Assuming acceleration is constant, the equations are:vf = vo + a*txf = xo + vo*t + (1/2) a*t^2vf^2 = vo^2 + 2*a*(xf-xo)d = (vf + vo)/2 * tOn the variables: f indicates final value, o indicates original or initial value.v = velocitya = accelerationx = positiond = distancet = time^ indicates an exponent (i.e. t^2 is t squared)* indicates multiplication

what is open kinematic chain?

Kinematic Self-Replicating Machines was created in 2004.

dynamic viscosity unit is poise and kinematic viscosity unit is stoke

Kinematic Self-Replicating Machines has 341 pages.

motion

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