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# The two ends of a train moving with a constant acceleration pass a certain pole with velocities $u$ and $v$. The velocity with which the middle point of the train passes the same pole is:-A. $\sqrt {\dfrac{{{u^2} + {v^2}}}{2}}$B. $\dfrac{{u + v}}{2}$C. $\sqrt {uv}$D. $\dfrac{{{u^2} + {v^2}}}{2}$ Verified
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Hint: Here we have to use the third equation of motion. The third equation of motion gives the final velocity of an object under uniform acceleration given the distance travelled and an initial velocity.
The third equation of motion is especially used to find the distance.
The third equation of motion is given mathematically as:
${v^2} - {u^2} = 2as$

In physics, equations of motion are classified as equations that characterise a physical system’s behaviour in terms of its motion as a function of time. For deriving components such as displacement, velocity, time and acceleration, there are three motion equations that can be used.
The equations of motion can also be said as the equations of kinematics.
The third equation does not give the time unlike the first two equations.
Given,
The two ends of a train moving with a constant acceleration pass a certain pole with velocities $u$ and $v$.
The velocity with which the middle point of the train passes the same pole is =?
We know that,
${v^2} - {u^2} = 2as$
$2as = {v^2} - {u^2}$ ...... (1)
Where $v =$ final velocity after travelling a distance $s$,
$u$ is the initial velocity,
$a$ is the acceleration.
Let the length of the train be given by $2as$.
Let the velocity of middle point of train at same point be $v'$, then
${\left( {v'} \right)^2} = {u^2} + 2a \times \dfrac{s} {2} = {u^2} + as$ ....... (2)
From (1) and (2) we get,
$v' = \sqrt {\dfrac{{{v^2} + {u^2}}}{2}}$

Hence, the velocity with which the middle point of the train passes the same pole is $\sqrt {\dfrac{{{v^2} + {u^2}}}{2}}$.
Therefore, option A is correct.

Note:Here when the train reaches the midpoint, we have to make sure that distance is half of the original distance otherwise we will not get the correct answer.