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Question: An engine of a train moving with uniform acceleration passes an electric pole with velocity u and th...

An engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last compartment with velocity v. The middle part of the train passes past the same pole with a velocity of:

A. u+v2 B. u2+v22 C. u2+v22 D. u2v22  A.{\text{ }}\dfrac{{u + v}}{2} \\\ B.{\text{ }}\dfrac{{{u^2} + {v^2}}}{2} \\\ C.{\text{ }}\sqrt {\dfrac{{{u^2} + {v^2}}}{2}} \\\ D.{\text{ }}\sqrt {\dfrac{{{u^2} - {v^2}}}{2}} \\\
Explanation

Solution

Hint- In order to solve this question, we will use the third law of motion, so first we will apply it using the distance between the first and last compartment in order to find the relation between initial and final speed and acceleration. Next we will use the same for half length of the train or the middle birth to find the relation between middle speed and the given speed using the acceleration.

Formula used- v2u2=2as{v^2} - {u^2} = 2as.

Complete step-by-step solution -
Let the length of train =l = l
When last compartment passes through pole then distance covered by first compartment is =l = l
According to the third equation of motion

v2u2=2as v2u2=2al a=v2u22l.........(1)  \because {v^2} - {u^2} = 2as \\\ \Rightarrow {v^2} - {u^2} = 2al \\\ \Rightarrow a = \dfrac{{{v^2} - {u^2}}}{{2l}}.........(1) \\\

Now, when middle part of the train passes, then the distance travelled by the first compartment is l2\dfrac{l}{2}
Using the third law of motion for this distance we get:

v2u2=2as vf2u2=2al2 vf2=u2+al.......(2)  \because {v^2} - {u^2} = 2as \\\ \Rightarrow v_f^2 - {u^2} = 2a\dfrac{l}{2} \\\ \Rightarrow v_f^2 = {u^2} + al.......(2) \\\

Now in order to find the given velocity let us substitute the value of acceleration from equation (1) into equation (2), here the acceleration is uniform so it will be the same for the last compartment and middle part.

vf2=u2+(v2u22l)l vf2=u2+(v2u22) vf2=2u2+v2u22 vf2=u2+v22 vf=u2+v22  \Rightarrow v_f^2 = {u^2} + \left( {\dfrac{{{v^2} - {u^2}}}{{2l}}} \right)l \\\ \Rightarrow v_f^2 = {u^2} + \left( {\dfrac{{{v^2} - {u^2}}}{2}} \right) \\\ \Rightarrow v_f^2 = \dfrac{{2{u^2} + {v^2} - {u^2}}}{2} \\\ \Rightarrow v_f^2 = \dfrac{{{u^2} + {v^2}}}{2} \\\ \Rightarrow {v_f} = \sqrt {\dfrac{{{u^2} + {v^2}}}{2}} \\\

Hence, the middle part of the train passes past the same pole with a velocity of u2+v22\sqrt {\dfrac{{{u^2} + {v^2}}}{2}}
So, the correct answer is option C.

Note- In order to solve such questions use the laws of motion but manipulate the distances in terms of length covered in terms of given length of train. Newton's motion laws are three physical laws which together laid the foundations for classical mechanics. They describe a body 's relationship with the forces acting on it, and the motion of the body in the response to those forces.