# Question ae78b

Sep 30, 2015

$\text{67 m}$

#### Explanation:

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

$\text{72 "color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("hr"))) * "1000 m"/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("hr"))))/"3600 s" = "20 m/s}$

SInce $\frac{\text{36 km/hr" = "72 km/hr}}{2}$, you get that

$\text{36 km/h" = "10 m/s}$

So, you know that it takes the train $\text{200 m}$ to reduce its speed from $20$ to $\text{10 m/s}$. This means that you can say

${v}^{2} = {v}_{0}^{2} - 2 \cdot a \cdot d \text{ }$, where

$a$ - the decelleration of the train.

Rearrange to solve for $a$

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

${\underbrace{{v}_{\text{final}}^{2}}}_{\textcolor{b l u e}{= 0}} = {v}^{2} - 2 \cdot a \cdot {d}_{x}$

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 $\text{200-m}$ distance

${d}_{x} = \textcolor{g r e e n}{\text{67 m}}$

You can test the result by using the total distance it travelled before being brought to rest and its initial speed, $\text{20 m/s}$

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

$400 = 2 \cdot 0.75 \cdot 267$

$400 = 400.5 \to$ close enough, if you take into account rounding

For ${d}_{x} = \text{66.7 m}$, you would get

$400 = 400.05$