- #1

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## Homework Statement

"When a high-speed passenger train traveling at 161 km/h rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance D = 676 m ahead (Fig.2-29). The locomotive is moving at 29.0 km/h. The engineer of the high-speed train immediately applies the brakes.

*What must be the magnitude of the resulting constant deceleration if a collision is to be just avoided?*"

## Homework Equations

##v_A=44.72 \frac{m}{s}##

##v_B=8.056 \frac{m}{s}##

##x_0=0m##

##x_1=676m##

##x_2 - x_1 = v_0t+\frac{1}{2}at^2##

##a_B = 0 \frac{m}{s^2}##

##x_2 = (8.056 \frac{m}{s})(t) + (676m)##

##x - x_0 = \frac{1}{2}(v_0+v)t##

*Answer as given by book: -0.994 m/s*

^{2}## The Attempt at a Solution

First, I used the last equation on the list, expressing ##x_2## in terms of t...

##(8.056 \frac{m}{s})(t) + (676m) - x_0= \frac{1}{2}(44.72\frac{m}{s})t##

since ##x_0=0\frac{m}{s^2}##

##(16.1\frac{m}{s})t+1352m=(44.72\frac{m}{s})t##

##t(16.1\frac{m}{s}-44.72\frac{m}{s}) = -1352m##

##t(-28.62\frac{m}{s})=-1352m##

##t=47.26s##

Then I plugged this time into the original distance equation to find the distance traveled by the locomotive plus 676:

##x_2 = (8.056 \frac{m}{s})(47.26) + (676m) = 1056.7m##

Then using this distance, I used the formula: ##x-x_0=v_0t+\frac{1}{2}at^2##, while substituting the appropriate values.

##1056.7m=(44.72 \frac{m}{s})(47.26s)+\frac{1}{2}a(47.26s)^2##

##2113.4m=4226.9344m+a(2233.51s^2)##

##-2113.5344m=a(2233.51s^2)##

##a=-0.946\frac{m}{s^2} ≠ -0.994\frac{m}{s^2}##

I have some confidence that I did this right, since the answers match up so closely. I was thinking the book had some kind of misprint, but I just wanted to see if I was doing this problem correctly or not, just to make sure.