Respuesta :
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Let's consider the given terms ~
- Initial velocity (u) = 100 m/s
- Acceleration (a) = - 10 m/s²
(Acceleration is negative bbecause train slows down)
- Final velocity (v) = 0 m/s
(final velocity is 0, because train stops)
Now, let's find the time taken (t) by using the first equation of motion ~
- [tex]v = u + at[/tex]
- [tex]0 = 100 + ( - 10 \times t)[/tex]
- [tex]0 = 100 - 10t[/tex]
- [tex]10t = 100[/tex]
- [tex]t = 100 \div 10[/tex]
- [tex]t = 10[/tex]
The train will take 10 secs to stop,
Now, let's find how much distance (s) it will cover before it stops using second equation of motion ~
- [tex]s = ut + \dfrac{1}{2} at {}^{2} [/tex]
- [tex]s = (100 \times 10) + ( \dfrac{1}{ 2} \times - 10 \times 10 \times 10)[/tex]
- [tex]s = 1000 + ( - 500)[/tex]
- [tex]s = 1000 - 500[/tex]
- [tex]500 \: \: m[/tex]
It will cover 500 meters before coming to rest ~