If the engine of a long train moving with constant acceleration crosses a tree with velocity ' u ' and the last compartment of the train crosses the same tree with velocity ' v ', then the velocity with which the middle compartment crosses the same tree is

Step-by-step solution

Using equation of motion, v ^2- u ^2=2 as let the length of train is L, on rearranging v^2-u^2=2 a La= v^2-u^2 2 L ....(1) The velocity with which middle compartment crosses pain, is given by, aligned & v_f^2-u^2=2 a S_f & v_f^2=u^2+2 a ( L 2 )=u^2+a L aligned Substitute "a" from equation (1) v_f^2=u^2+ ( v^2-u^2 2 L ) L v _ f = v ^2+ u ^2 2