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# Final Velocity (from constant a)

MichaelBartmess.Final Velocity (from constant a)

This equation computes the square of the final velocity that a body would achieve after traveling in a straight line some distance at constant acceleration. This an illustrative step in calculating the actual `v_f` based on acceleration and time. See the derivation below.

The remaining step, to take the square root of both sides of this equation happens in the sister equation:

## INPUTS

- `x_i` - the initial displacement
- `x_f` - the final displacement
**a**- the constant acceleration- `V_0` - the initial velocity

## DERIVATION

Since acceleration is constant, we know that the final velocity is the sum of the initial velocity and the velocity increase due to the acceleration. In other words:

[1] `V_f = V_i + a * t`

We also know that the distance traveled, d, is the sum of the distance the object would travel at its starting velocity, `V_i`, plus the distance it would travel while increasing velocity from `V_i` to `V_f`:

[2] `D = (V_i * t) + (1/2 * (V_f - V_i) * t)`

[3] `D = t * (V_i + 1/2 * V_f - 1/2 * V_i)`

[4] `D = t * 1/2 (V_i + V_f)`

[5] `=> t = (2 * D) / (V_i + V_f)`

Substituting [5} into [1]:

[6] `V_f = V_i + a * ((2 * D) / (V_i + V_f))`

Multiplying both sides by '(V_i + V_f)`:

[7] `V_i *V_f + V_f^2 = V_i^2 + V_i * V_f + 2*A*D`

Cancelling term `V_i* V_f`:

[8] `V_f^2 = V_i^2 + 2*a*D`, where `D = x_f - x_0`

[9] **`V_f^2 = V_i^2 + 2*a*(x_f - x_0)`**

This equation [9] computes the resultant **`V_f^2**`, which is not useful in most cases, so we want to get the square root of this resultant:

[10] **`V_f = sqrt(V_i^2 + 2*a*(x_f - x_0))`**

## EXTERNAL LINKS

Khan Academy's Average velocity for constant acceleration

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