Solving for Final Velocity II

Solving for Final Velocity When Velocity is not Constant (a ≠ 0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve \(v = v_0 + at\) for \(t,\) we get

\(t = \cfrac{v - v_0}{a}.\)

Substituting this and \(\bar{v} = \frac{v_0 + v}{2}\) into \(x = x_0 + \bar{v} t,\) we get

Example on Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in the example from the previous lesson without using information about time. Recall that dragsters can achieve average accelerations of 26.0 m/s². Also, recall that \(x = 402\text{ m}.\)

Strategy

Draw a sketch.

The equation \(v^2 = v_0^2 + 2a(x \; - \; x_0)\) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that \(v_0 = 0,\) since the dragster starts from rest. Then we note that \(x \; - \; x_0 = 402\text{ m}\) (this was the answer in the example). Finally, the average acceleration was given to be \(a = 26.0\mathrm{\; m/s^2}.\)

2. Plug the knowns into the equation \(v^2 = v_0^2 + 2a(x \; - \; x_0)\) and solve for \(v.\)

\(v^2 = 0 + 2 \left ( 26.0\mathrm{\; m/s^2} \right ) (402\text{ m}).\)

Thus

\(v^2 = 2.09 \times 10^4\mathrm{\; m^2/s^2}.\)

To get \(v,\) we take the square root:

\(v = \sqrt{2.09 \times 10^4\mathrm{\; m^2/s^2}} = 145\text{ m/s}.\)

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation \(v^2 = v_0^2 + 2a(x \; - \; x_0)\) can produce further insights into the general relationships among physical quantities:

• The final velocity depends on how large the acceleration is and the distance over which it acts
• For a fixed deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)