JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 9)
An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :
$${{u + v} \over 2}$$
$$\sqrt {{{{v^2} - {u^2}} \over 2}} $$
$${{v - u} \over 2}$$
$$\sqrt {{{{v^2} + {u^2}} \over 2}} $$
Explanation
_25th_February_Morning_Shift_en_9_2.png)
Let initial speed of train u. When midpoint of the train reach the signal post it's velocity becomes v0.
$$ \therefore $$ $$v_0^2 = {u^2} + 2as$$ .......(1)
When train passes the signal post completely it's velocity becomes v.
$$ \therefore $$ $${v^2} = v_0^2 + 2as$$ ......(2)
Subtracting (2) from (1) we get,
$$v_0^2 - {v^2} = {u^2} - v_0^2$$
$$ \Rightarrow v_0^2 + v_0^2 = {u^2} + {v^2}$$
$$ \Rightarrow {v_0} = \sqrt {{{{u^2} + {v^2}} \over 2}} $$
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