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v1 = initial velocity

v2 = final velocity

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usually v1 is initial speed and v2 final speed

Q: In the acceleration equation what do v1 and v2 represent?

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Delta "T"=V2-V1 ---- A

1. If V1 and V2 be two vectors at 900 from each other, having magnitudes of 10 and 20 units each, what will be the value of V1.V2 ?

Yes, a unit vector can have negative component since a unit vector has same magnitude and direction as a negative unit vector. Here is the general work out of the problem: Let |v| be the norm of (v1, v2). Then, the unit vector is (v1/|v|, v2/|v|). Determine the "modulus" or the norm |(v1/|v|, v2/|v|)| to get 1, which is the new norm. If we determine the norm of |(-v1/|v|, -v2/|v|)|, we still have the same norm 1.

Calculate percentage changefrom V1 = 3624 to V2 = 8236[ ((V2 - V1) / |V1|) * 100 ]= ((8236 - 3624) / |3624|) * 100= (4612 / 3624) * 100= 1.272627 * 100= 127.2627% change= 127.26% increase.

A Compound Graph is an extension of a standard graph. Let G be a graph, G=(V,E) where V is a set of vertices and E is a set of edges, that is e = (v1, v2) in V2 A compound graph C is defined by a tree T=(V,F) where V is the same set as G and F are tree edges f=(v1,v2) in V2. C=(G,T) where G=(V,E) and T=(V,F) Furthermore, C has two additional constraints: e=(v1,v2) in E implies: 1) v1 is not on the path of v2 to the root of T AND 2) v2 is not on the path of v1 to the root of T. Intuitively, T defines a hierarchy. All the vertices sharing the same parent in T are in the same "group". The constraints state that you cannot have an edge connecting a vertex to one of its parent in the hierarchy.

Related questions

The equations of motion that relate velocity, distance, time and acceleration for the specific case of "constant acceleration" can be written as follow, acceleration a = (v2 - v1)/t from which v2 = v1 + at The distance covered during t time d = vav x t, where vav refers to average velocity in the process from v1 to v2. For the case of constant acceleration vav = (v1 + v2)/2. Substituting in d we get d = (v1 + v2)/2 x t from which, v2 = 2d/t - v1 If we take the constant acceleration to be zero, a = 0, you can see that the second equation we wrote becomes, v2 = v1 (There is no acceleration), so our equation for the distance d becomes, d = v1 x t = v2 x t

( | V1 - V2 | / ((V1 + V2)/2) ) * 100

The equation for average acceleration is: average acceleration = change in velocity / change in time.

Okay here is what you want to do. You rearrange your equation of V1=a/d so that it looks like this --> v2= dxa and that is how you get your answer

Delta "T"=V2-V1 ---- A

v0=v1+v2

(p1/v1) = (p2/v2)For Apex (P1 N1)= (P2N2 )

To solve Boyle's Law equation for V2, first write the equation as P1V1 = P2V2. Then rearrange it to isolate V2 on one side, dividing both sides by P2 to solve for V2, which will be V2 = (P1 * V1) / P2.

A1V1=A2V2 or V2=(A1/A2)(V1)

The formula to determine acceleration is acceleration = change in velocity / time taken. It can also be written as a = (v2 - v1) / t, where a is acceleration, v1 is the initial velocity, v2 is the final velocity, and t is the time taken.

Vo=(R2/R1)(V2-V1)

AnswerAcceleration = v2 - v1.........................timeAverage Acceleration requires you to average the the initial velocity of your trials and the ending velocity of your trials. You must also find the average for your time trials. Once you achieve these numbers, just plug them into the above formula and you can solve for average acceleration.