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?

Question

Amiyah Price

Physics

8 February, 04:26

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?

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Answers (1)

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Ciara Miller · Answer 1

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