The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. the freight train is traveling at 15.0 m/s in the same direction as the passenger train. the engineer of the passenger train immediatly applies the brakes causing a constant acceleration of .100m/s^2 in a direction opposite to the trains velocity, while the freight train continues with constant speed. will there be a collision?

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2  + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

a and b sometimes have a positive value