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Omid
Sep15-04, 11:41 AM
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.

HallsofIvy
Sep15-04, 12:16 PM
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.

Omid
Sep15-04, 12:34 PM
Ok.
What if two accelerations were equal ?
Thank you

BobG
Sep15-04, 12:59 PM
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.

Chronos
Sep15-04, 03:55 PM
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.