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

1 Answer
Oct 4, 2017

Answer:

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