How do I find the distance between two trains heading towards each other?

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Two trains, initially 250 meters apart, are decelerating towards each other, with Train A traveling at 96 km/h and Train B at 110 km/h. The calculations show that Train A stops after approximately 6.67 seconds, while Train B continues to decelerate for about 10.18 seconds. The initial approach incorrectly assumed a constant relative deceleration, leading to a distance calculation of 233.9 meters, which is less than the required 250 meters for a collision. The correct method involves treating each train's deceleration separately and summing the distances traveled before they stop. The discussion emphasizes the importance of considering the different deceleration times for each train to accurately determine if a collision occurs.
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Two trains heading straight for each other on the same track are 250 m apart when their engineers see each other and hit the brakes. The A, heading west at 96 km/h, slows down, accelerating at an average of -4 m/s^2, while the B, traveling at 110 km/h, slows down, accelerating at an average of -3 m/s^2. Will they collide?


I solved it this way :
a_T = The resultant of two accelerations = -7
V_A = 96 km/h
V_B = 110 km/h
V_T = V_B + V_A = 206 km/h = 57.22 m/s
And considered (V_T)_final = 0.
Plugging into this formula : S = - [(V_T)^2 ] / 2 a_T; S = 233.9. So two trains won't collide.
My answer for S is not the same as the one in the solutions manual, 244.5 m. Why is my answer 233.9 ?
The book has solved that problem in a different way, and I can do it so, but I'm interested to know why my answer is wrong.
 
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The two trains do not decelerate for the same length of time.

The train moving at 96 kph = 26.67 m/s decelerates at -4 m/s2 for 26.67/4= 6.67 seconds at which time it is stopped. The train moving at 110 kph= 30.5 m/s decelerates at -3 m/s2 for 30.5/3= 10.18 seconds. The two trains will have a "relative deceleration" of -7 m/s2 for the first 6.67 seconds but after that the first train has stopped and the "relative deceleration" is only -3 m/s2. You could do the problem by considering those to time intervals separately but it is easier to treat the two trains separately- which is what I presume you book did- that's what I did and got the same answer as your book.
 
Thank you

Ok.
What if two accelerations were equal ?
Thank you
 
If the accelerations were equal, your approach would be fine. In fact, if you substituted a force that could send the train in reverse (thrusters or spin the wheels backward, for example), your approach would be fine, provided you subtracted your answer from the 250 m to find out how close the trains would come to each other.

Your answer was wrong because of the special case given - brakes can only slow you down to 0, they can't send you into reverse.
 
Break problem into two parts. First solve for a final velocity of 0 for each train using the formula Vf = Vi + AT and solve for T for each train. Next, plug T for each train into the distance formula D = AT^2/2. Next add up the distance each train travels.
 
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