# Question #ae78b

##### 1 Answer

#### Explanation:

To make the calculations esier, convert the speeds of the train from *kilometers per hour* to *meters per second*

#"72 "color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("hr"))) * "1000 m"/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("hr"))))/"3600 s" = "20 m/s"#

SInce

#"36 km/h" = "10 m/s"#

So, you know that it takes the train

#v^2 = v_0^2 - 2 * a * d" "# , where

Rearrange to solve for

#a = (v_0^2 - v^2)/(2 * d) = ((20""^2 - 10""^2)"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 200color(red)(cancel(color(black)("m")))) = "0.75 m/s"""^2#

Now, I assume that you meant to say *how much further* will the train go until it's brought to rest. Well, if the train decelerates at constant rate, and provided, of course, that the speed of the train is **zero** once it comes to a rest, you can say that

#underbrace(v_"final"^2)_(color(blue)(=0)) = v^2 - 2 * a * d_x#

This means that it travels *for an additional*

#d_x = v^2/(2 * g) = (10""^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 0.75color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = "66.7 m"#

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the

#d_x = color(green)("67 m")#

You can test the result by using the **total distance** it travelled before being brought to rest and its initial speed,

#0^2 = 20""^2 - 2 * 0.75 * (200 + 67)#

#400 = 2 * 0.75 * 267#

#400 = 400.5 -># close enough, if you take into account rounding

For

#400 = 400.05#