A train is moving with constant velocity. Suddenly its last compartmen...
Problem:
A train is moving with constant velocity. Suddenly its last compartment gets detached and stops after traveling d distance and uniform retardation. If the remaining train continued to move with uniform velocity, find the distance before which the train stops in terms of d?
Solution:
Let the length of the train be L and the initial velocity of the train be u. After the last compartment gets detached, the remaining train continues to move with the same uniform velocity u. Let the distance covered by the remaining train before coming to rest be x.
Step 1: Finding the time taken by the last compartment to come to rest
Using the first equation of motion,
v = u + at
where v is the final velocity of the last compartment, which is 0, and a is the retardation.
Therefore,
0 = u + at
t = -u/a
Step 2: Finding the distance covered by the last compartment before coming to rest
Using the third equation of motion,
s = ut + 1/2at^2
Substituting the value of t from step 1, we get
s = ud/(-2u/a) = -ad/2
Step 3: Finding the distance covered by the remaining train before coming to rest
The remaining train is moving with uniform velocity u. Therefore, using the third equation of motion,
s = vt
where v is the velocity of the remaining train and t is the time taken to come to rest.
Substituting the value of t from step 1, we get
s = u(-u/a) = -u^2/a
Step 4: Finding the total distance covered by the train before coming to rest
The total distance covered by the train before coming to rest is the sum of the distances covered by the last compartment and the remaining train.
Total distance = -ad/2 - u^2/a
On simplifying,
Total distance = (-au^2 - 2d)/2a
Step 5: Final Answer
Therefore, the distance before which the train stops is (-au^2 - 2d)/2a in terms of d.