# Distance from a Changing Velocity

## Key Questions

$\Delta v = \overline{a} \times \Delta t$

#### Explanation:

This is the equation for acceleration:
$\overline{a} = \frac{\Delta v}{\Delta t}$

Let's solve for $\Delta v$, which means change in velocity

$\overline{a} = \frac{\Delta v}{\Delta t}$

multiply by $\Delta t$ on both sides

$\Delta v = \overline{a} \times \Delta t$

Now we have the equation to find the change in velocity

• Rate of change of velocity is called acceleration. When this rate of change is constant, it is called uniform acceleration. Like the change of velocity of an object falling freely under gravity. Its velocity increases by 9.8 m every second. This acceleration is denoted as 9.8m/s^2

The limit to find the velocity represents the real velocity, whereas without the limit one finds the average velocity.

#### Explanation:

The physics relationship of them using averages is:

$u = \frac{s}{t}$

Where $u$ is the velocity, $s$ is the distance traveled and $t$ is the time. The longer the time, the more accurate the average speed can be calculated.

However, although the runner could have a velocity of $5 \frac{m}{s}$ those could be an average of $3 \frac{m}{s}$ and $7 \frac{m}{s}$ or a parameter of infinite velocities during the time period. Therefore, since increasing time makes velocity "more average" decreasing time makes velocity "less average" therefore more precise. The smallest value that time could take would be 0, but that would nulify the denominator. Therefore, one uses the limit as $t$ tends to, but never approaches, 0.

$u = {\lim}_{t \to 0} \left(\frac{s}{t}\right)$

• If the velocity is changing at a constant rate, meaning that the acceleration is constant, we may derive the formula as follows.

The velocity $v$ varies linearly with time and is given by the relation $v \left(t\right) = {v}_{0} + a t$ where ${v}_{0}$ is the initial velocity and $a$ is the acceleration.

We know that distance is the product of average velocity and time, and the average velocity is the average of the initial and final velocity. In mathematical terms,

${v}_{a v g} = \setminus \frac{{v}_{0} + v}{2} = \setminus \frac{{v}_{0} + \left({v}_{0} + a t\right)}{2} = {v}_{0} + \setminus \frac{a t}{2}$

Simply multiply that by time $t$ to get the distance $s$.

$s = \left({v}_{0} + \frac{a t}{2}\right) \cdot t = {v}_{0} t + \frac{1}{2} a {t}^{2}$