# How to find deceleration to stop collision with moving body?

1. Feb 21, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"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?"

2. Relevant 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

3. 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.

2. Feb 21, 2016

### ehild

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

3. Feb 21, 2016

### 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.

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

4. Feb 21, 2016

### haruspex

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?

5. Feb 21, 2016

### 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?

6. Feb 21, 2016

### haruspex

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

7. Feb 21, 2016

### Eclair_de_XII

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

Last edited: Feb 21, 2016