a train A moving at 38m/s is moving to the right and is 220 meters away from ano
ID: 2170869 • Letter: A
Question
a train A moving at 38m/s is moving to the right and is 220 meters away from another train B moving at 44m/s to the left. Train A moving to the right decelerates at a rate of "a" m/s^2 and train B moving to the left decelerates at a rate of 4m/s^2.a) What should be the minimum deceleration of train A be so that after it stops, the passengers have 5 seconds to evacuate the train? [assume that train A will stop and not move backwards]
b) Where will the crash take place and how fast is train B moving at that time?
Explanation / Answer
A) Train A A = a m/s^2 Vi = 38 m/s Xi = 0 m Train B A = 4 m/s^2 Vi = -44 m/s Xi = 220 m So we want train A to stop 5 seconds before the collision with train B occurs. If we let t = c be the time of collision, then it takes c-5 seconds for Train A to stop. Since train A doesn't move between t= c-5 and T=c, we know that the displacement of train A at t=c-5 is equal to the displacement of train B at t=c. The equation for train A is Xf = Xi + Vit + (1/2)At^2 Xf = 0 +38(c-5) + (1/2)a(c-5)^2 The equation for train B is Xf = 220 - 44c + (1/2)(4)c^2 Now we set them equal: 0 +38(c-5) + (1/2)a(c-5)^2 = 220 - 44c + (1/2)(4)c^2 We have 1 too many variables here so let's see if we can get c in terms of a: For train A: Vf = 38 + at 0 = 38 + a(c-5) a = -38/(c-5) Great. Let's revisit the equation. 0 +38(c-5) + (1/2)a(c-5)^2 = 220 - 44c + (1/2)(4)c^2 0 +38(c-5) + (1/2)(-38/(c-5))(c-5)^2 = 220 - 44c + (1/2)(4)c^2 Solving for c gives us c= 6.23 or c=25.26 We can plug these into the equation for train B to see which makes sense: Using c=25.26 for gives us a final position of B that is to the right of where it started, so that isn't right. So c= 6.23 a = -38/(c-5) = -38/(6.23-5) = -30.89 m/s^2 B) We plug c into either motion equation to get: 23.51 m (this is to the right of where train A started) Again using c and the equation for velocity of train B: Vf = Vi + at Vf = -44 + 4(6.23) = -19.08 m/s (the negative just indicates that it is moving to the left... as it should be)
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