Final Velocity (from constant a)

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Equation / Last modified by KurtHeckman on 2016/01/11 21:25
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distance at constant acceleration.pngThis 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))`

Khan Academy's Average velocity for constant acceleration

 

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