# An object moving at 8.293485234671349xx10^23094m/s stops after 12398475s what was the object's acceleration?

## please calculate the object's acceleration

Oct 4, 2017

What is a tremedous speed and acceleration! We are in the outer space! ($a = - 6.6891172 \times {10}^{23087} m \text{/} {s}^{2}$)

#### Explanation:

In a constant acceleration linear motion, velocity($v$), time($t$) and acceleration($a$) follows the three formula.

$v = {v}_{0} + a t$ (1)
$x = {v}_{0} t + \frac{1}{2} a {t}^{2}$ (2)
${v}^{2} - {v}_{0}^{2} = 2 a x$ (3)

Here, ${v}_{0}$ is the initial speed (speed at $t = 0$) and $x$ is the displacement (how long the obeject has moved).

For this question, what you know are:
${v}_{0} = 8.293485234671349 \times {10}^{23094} \text{ "m"/} s$
$v = 0 \text{ "m"/} s$
$t = 12398475 s$
and what you want to know: $a \text{ "m"/} {s}^{2}$.

Use the formula (1), you can see
$a = \frac{v - {v}_{0}}{t}$
$= \frac{0 - 8.293485234671349 \times {10}^{23094} m \text{/} s}{12398475 s}$
$= - 6.6891172 \times {10}^{23087} m \text{/} {s}^{2}$

$\textcolor{red}{\text{However, this never happens!}}$
Albert Einstein has proved that nothing travels faster than light
($c = 3.0 \times {10}^{8} m \text{/} s$) in his special relativity.
https://en.wikipedia.org/wiki/Special_relativity