# What is the formula for time from a changing velocity?

Jan 5, 2016

$t = \frac{u - {u}_{0}}{a}$

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$ (Need to solve quadratic)

#### Explanation:

Via changing velocity I pressume you mean an object that accelerates or decelerates.

If acceleration is constant

If you have initial and final speed:

a=(Δu)/(Δt)

$a = \frac{u - {u}_{0}}{t - {t}_{0}}$

Usually ${t}_{0} = 0$, so:

$t = \frac{u - {u}_{0}}{a}$

If the above method does not work because you are missing some values, you can use the equation below. The distance traveled $s$ can be given from:

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$

where ${u}_{0}$ is the initial speed
$t$ is the the time
$a$ is the acceleration (note this value is negative if the case is a deceleration)

Therefore, if you know the distance, initial speed and acceleration you can find the time by solving the quadratic equation that is formed. However, if acceleration if not given, you will need the final speed of the object $u$ and can use the formula:

$u = {u}_{0} + a t$

$u - {u}_{0} = a t$

$a = \frac{u - {u}_{0}}{t}$

and substitute to the distance equation, making it:

$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \frac{u - {u}_{0}}{t} \cdot {t}^{2}$

$s = {u}_{0} \cdot t + \frac{1}{2} \cdot \left(u - {u}_{0}\right) \cdot t$

Factor $t$:

$s = t \cdot \left({u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)\right)$

$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$

So you got 2 equations. Pick one of them, which will help you solve with the data you are given:

$s = {u}_{0} \cdot t + \frac{1}{2} a {t}^{2}$

$t = \frac{s}{{u}_{0} + \frac{1}{2} \cdot \left(u - {u}_{0}\right)}$

Below are two other cases where acceleration is not constant. FEEL FREE TO IGNORE THEM if acceleration in your case is constant, since you placed it in the Precalculus category and the below contain calculus.

If acceleration is a function of time $a = f \left(t\right)$

The definition of acceleration:

$a \left(t\right) = \frac{\mathrm{du}}{\mathrm{dt}}$

$a \left(t\right) \mathrm{dt} = \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = {\int}_{{u}_{0}}^{u} \mathrm{du}$

${\int}_{0}^{t} a \left(t\right) \mathrm{dt} = u - {u}_{0}$

$u = {u}_{0} + {\int}_{0}^{t} a \left(t\right) \mathrm{dt}$

If you still don't have enough to solve, that means you have to go to distance. Just use the definition of speed and move on, as if I analyze it further it will only confuse you:

$u \left(t\right) = \frac{\mathrm{ds}}{\mathrm{dt}}$

The second part of this equation means integrading acceleration with respect to time. Doing that gives an equation with only $t$ as the unknown value.

If acceleration is a function of speed $a = f \left(u\right)$

The definition of acceleration:

$a \left(u\right) = \frac{\mathrm{du}}{\mathrm{dt}}$

$\mathrm{dt} = \frac{\mathrm{du}}{a \left(u\right)}$

${\int}_{0}^{t} \mathrm{dt} = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$

$t - 0 = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$

$t = {\int}_{{u}_{0}}^{u} \frac{\mathrm{du}}{a \left(u\right)}$