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Question: An express train is moving with a velocity \(\mathop v\nolimits_1 \), Its driver finds another train...

An express train is moving with a velocity v1\mathop v\nolimits_1 , Its driver finds another train is moving on the same track in the same direction with velocity v2\mathop v\nolimits_2 s. To escape collision. the driver applies retardation on the train. The minimum time of escaping collision will be:-
(A) v1−v2a\dfrac{{\mathop v\nolimits_1 - \mathop v\nolimits_2 }}{a}
(B) v12−v222\dfrac{{\mathop {\mathop v\nolimits_1 }\nolimits^2 - \mathop {\mathop v\nolimits_2 }\nolimits^2 }}{2}
(C) None
(D) Both

Explanation

Solution

Hint Collision is defined as a process when two bodies collide with each other and sometimes deformation occurs in their shape and sometimes deformation gets reformed. So, collisions are classified as elastic collision and inelastic collision. During collision total momentum of the system remains conserved.

Complete step by step answer:
As both the trains are moving and they become frame of reference for each other, we will first calculate relative velocity between them :-
Initial relative velocity will be =v1−v2= \mathop v\nolimits_1 - \mathop v\nolimits_2
Now as the driver applies break and train gets retarded so both the train will move with same velocity that is =v1−v2= \mathop v\nolimits_1 - \mathop v\nolimits_2
So now final relative velocity between them will be v1−v2=0\mathop v\nolimits_1 - \mathop v\nolimits_2 = 0
Applying the equation of motion :- v = u + atv{\text{ }} = {\text{ }}u{\text{ }} + {\text{ }}at
Now final relative velocity = 0
Initial relative velocity = u = v1 − v2 u{\text{ }} = {\text{ }}\mathop v\nolimits_1 {\text{ }}-{\text{ }}\mathop v\nolimits_2 {\text{ }}
Now putting values
0=(v1 - v2) + at 0 = (\mathop v\nolimits_1 {\text{ - }}\mathop v\nolimits_2 {\text{) + at }}
As body is retarded then we will put negative sign with a
0=(v1 - v2) - at 0 = (\mathop v\nolimits_1 {\text{ - }}\mathop v\nolimits_2 {\text{) - at }}
(v1 - v2) = at (\mathop v\nolimits_1 {\text{ - }}\mathop v\nolimits_2 {\text{) = at }}
t=v1a−v2at = \dfrac{{\mathop v\nolimits_1 }}{a} - \dfrac{{\mathop v\nolimits_2 }}{a}

Now we will match the calculated value with the given options: -
v1−v2a\dfrac{{\mathop v\nolimits_1 - \mathop v\nolimits_2 }}{a}As this option exactly matches with the calculated value. Thus, this option is correct.
v12−v222\dfrac{{\mathop {\mathop v\nolimits_1 }\nolimits^2 - \mathop {\mathop v\nolimits_2 }\nolimits^2 }}{2}: This option does not match with the calculated value. Hence this option is not correct.
C. None: - as option is exactly correct. This option is not correct.
D. Both: - as the b option is not correct. Thus, this option is not correct.

Thus, the required answer is option A v1−v2a\dfrac{{\mathop v\nolimits_1 - \mathop v\nolimits_2 }}{a}

Note A collision is an isolated event in which two or more colliding bodies exert strong forces on each other for a relatively short time. For a collision to take place, the actual physical contact is not necessary. Total linear momentum is conserved in all collisions.