How to find deceleration to stop collision with moving body?

[Eclair_de_XII]

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/s²

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.

 

[ehild]

The train does not need to stop to avoid collision with the other one.

 

[Eclair_de_XII]

I get it. So it just has to slow down to the point that its velocity equals that of the locomotive it is to avoid contact with.

$(8.056\frac{m}{s})t+676=\frac{1}{2}(44.72\frac{m}{s}+8.056\frac{m}{s})t$
$(16.11\frac{m}{s})+1352m=(52.78\frac{m}{s})t$
$t=36.873s$
$x_2=(8.056\frac{m}{s})(36.873s)+676m=973.03m$
$973.03=(44.72\frac{m}{s})(36.873s)+\frac{1}{2}a(36.873s)^2$
$1946.061m=3297.92m+a(1359.62s^2)$
$-1351.8594m=a(1359.62s^2)$
$a=-0.994\frac{m}{s^2}$

 

[haruspex]

I get it. So it just has to slow down to the point that its velocity equals that of the locomotive it is to avoid contact with.

$(8.056\frac{m}{s})t+676=\frac{1}{2}(44.72\frac{m}{s}+8.056\frac{m}{s})t$
$(16.11\frac{m}{s})+1352m=(52.78\frac{m}{s})t$
$t=36.873s$
$x_2=(8.056\frac{m}{s})(36.873s)+676m=973.03m$
$973.03=(44.72\frac{m}{s})(36.873s)+\frac{1}{2}a(36.873s)^2$
$1946.061m=3297.92m+a(1359.62s^2)$
$-1351.8594m=a(1359.62s^2)$
$a=-0.994\frac{m}{s^2}$

Well done, but there is an easier way. Consider the motions relative to the slow locomotive. That one is stationary; how fast is the other approaching, at first? Given the initial relative speed, the final relative speed (0) and the initial separation, what equation will give you the acceleration directly?

 

[Eclair_de_XII]

Relative speed, huh? So the speed of the faster locomotive relative to the slower one. I tried fooling around with the numbers, subtracting the speed of the slower train from the faster train.

$v=44.72\frac{m}{s}-8.05\frac{m}{s} = 36.67\frac{m}{s}$
$v_0=0\frac{m}{s}$
$x=0m$
$x_0=676m$
$v^2=v_0^2+2a(x-x_0)$
$(36.67\frac{m}{s})^2=2a(-676m)$
$1344.44\frac{m^2}{s^2}=2a(-676m)$
$a=-0.944116\frac{m}{s^2}$

Is that right?

 

[haruspex]

Relative speed, huh? So the speed of the faster locomotive relative to the slower one. I tried fooling around with the numbers, subtracting the speed of the slower train from the faster train.

$v=44.72\frac{m}{s}-8.05\frac{m}{s} = 36.67\frac{m}{s}$
$v_0=0\frac{m}{s}$
$x=0m$
$x_0=676m$
$v^2=v_0^2+2a(x-x_0)$
$(36.67\frac{m}{s})^2=2a(-676m)$
$1344.44\frac{m^2}{s^2}=2a(-676m)$
$a=-0.944116\frac{m}{s^2}$

Is that right?

That's the right method, but using it I got -0.994.

 

[Eclair_de_XII]

Oh, I made a typo; it was -0.9944116.