Homework SourseYoure an investigator for the National Transportation Safety
You\'re an investigator for the National Transportation Safety Board, examining a subway accident in which a train going at 86 km/hcollided with a slower train traveling in the same direction at 23 km/h . Your job is to determine the relative speed of the collision, to help establish new crash standards. The faster train\'s \"black box\" shows that it began negatively accelerating at 1.8 m/s2 when it was 52 m from the slower train, while the slower train continued at constant speed.
Solution
>> Now, Initially,
Speed of Faster Train, U1 = 86 km/h = 23.889m/s
Speed of Slower train, U2 = 23 km/h = 6.3889 m/s
Distance between both trains = 52 m
Deacceleration of Fast Train, a = - 1.8 m/s2
>> Now, Let\'s assume that it takes time \'t\' from initial position to crash
>> Now, in this time \'t\', distance moved by slower train, S2 = U2*t = 6.3889*t
>> and, Distance Covered by Faster train, S1 = (U1)t + (1/2)*a*t2 [ by Newton\'s 2nd Equation of Motion ]
=> S1 = 23.889*t - 0.9*t2
>> Now, for Crash, Distance Covered by Faster train = Original Seperation + Distance Covered by Slower Train
=> S1 = 52 + S2
=> 23.889*t - 0.9*t2 = 52 + 6.3889*t
Solving it,
t = 3.661 sec
>> So, Velocity of Faster Train at timeof Crash, V1 = U1 + a*t [ by Newton\'s 1st Equation of Motion ]
=> V1 = 23.889 -1.8*3.661
=> V1 = 17.299 m/s
>> As, Velocity of Slower Train is Constant
=> V2 = U2 = 6.3889 m/s
So, Relative Velocity of Crash = V1 - V2
= 17.299 - 6.3889
= 10.910 m/s
= 39.277 Km/h ...ANSWER.....
