Train1 traveling 25 m/s spots another train 200 m in front. Train2 traveling 15 m/s. train1 breaks -0.1 m/s^2.

Question

Running on the same tracks in the same direction. Train1 applies breaks causing a -0.1 m/s^2 constant acceleration. At what point do the trains collide?

How did you get it?

At what position. In meters.

Answer 1

(position train 2) = 200+ 15t
(position train 1) = 0+ 25t- .1t²/2 = 25t- .05t²

There is a collision when (position train 1) = (position train 2),
200+ 15t = 25t- 0.05t²
.05t²- 10t+ 200 = 0
t = (10+ sq. root of 60)/.1 or (10- sq. root of 60)/.1
t = 177.5 sec or 22.54 sec

Let's ignore the bigger value since the trains are going to crash into each other only once.

The position of both trains at t = 22.54 sec is then 200+ 15(22.54) = 538.1 m. I came to this value by plugging back the value of t, 22.54 sec, into the first formula for train 2's position.

Answer 2

omg f*ing trains.
distance train 1=25t-1/2(.1)t^2
distance train 2=15t
since d1+d2 is 200
200=25t+15t-1/2(.1)t^2
then you solve for t
and plug t into either original equation for a distance
im really tired so im not doing this