# Determine whether there is a train collision

Two trains, one traveling at 60 miles/hr and the other at 80 miles/hr, are headed toward one another on a straight level track. When they are 2.0 miles apart, both engineers simultaneously see the other's train and apply their brakes. If the brakes decelerate each train at the rate of 3.0 ft/sec^2, determine whether there is a collision.

Ok, so is it correct to say that if $d > \frac{(v_{1}-v_{2})^{2}}{2a}$ there will be no collision, and if $d < \frac{(v_{1}-v_{2})^{2}}{2a}$ there will be a collision (d is distance, v is velocity, and a is acceleration). When I do this, I get that there will be a collision, but the correct answer is that there will be no collision. What am I doing wrong?

Thanks

StatusX
Homework Helper
You need to use two equations and see how far each train gets, and then see whether the sum is greater or less than 2 miles. The reason you can't get away with one equation is because once the 60 mph train stops, it doesn't start going backwards. It stops accelerating, but the 80 mph train continues slowing down. Plus, if you were going to do it that way, you would use v1+v2 in your equation, and 6 ft/sec^2 as the acceleration, since both trains are accelerating.

would I use $x = x{0} + v_{x}_{0}t + \frac{1}{2}(v_{x}_{0} + v_{x})$?

StatusX
Homework Helper
I think your last term should be 1/2 a t^2, and remember you need to find the t where each train stops and plug that in to get x. You don't need an x_0, you just need to know if the total distance the trains travel once they start braking is more or less than 2 miles.

So $v_{x} = v_{x}_{0} + a_{x}t$. I got $t = 20, t = 26.66$ for 60 mph and 80 mph respectively. After plugging these times in $x = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}$ I got $1066.1334 + 600 = 1666.1334 ft$ which is less than two miles. Is this correct?

StatusX
Homework Helper