Question

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

please calculate the object's acceleration

Answer 1

What is a tremedous speed and acceleration! We are in the outer space! (`a=-6.6891172xx10^23087 m"/"s^2`)

Explanation:

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

`v=v_0+at` (1)
`x=v_0t+1/2at^2` (2)
`v^2-v_0^2=2ax` (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=8.293485234671349xx10^23094" "m"/"s`
`v=0" "m"/"s`
`t=12398475 s`
and what you want to know: `a" "m"/"s^2`.

Use the formula (1), you can see
`a=(v-v_0)/t`
`=(0-8.293485234671349xx10^23094 m"/"s)/(12398475 s)`
`=-6.6891172xx10^23087 m"/"s^2`

`color(red)"However, this never happens!"`
Albert Einstein has proved that nothing travels faster than light
(`c=3.0xx10^8 m"/"s`) in his special relativity.
https://en.wikipedia.org/wiki/Special_relativity