Question

plz solve the problem___the answer is options 3)

Answer 1

Let 2L be the length of the train and a be it’s acceleration. Suppose, v is velocity of the mid point when it passes by the said point.

For motion from one end to mid point, we have,

v^2-u^2=2aL………………….(1).

For motion of other half, we have,

9u^2-v^2=2aL………………(2)

From (1) and(2), we get,

9u^2-v^2=v^2-u^2. This gives,

2v^2=10u^2

Or

v=sqrt(5)

Answer 2

Heya!

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Answer is,

Note -

S = Distance
v = final velocity
u = initial velocity
a = acceleration

S = (v² - u²)/2a

Final velocity given = 3u
Initial velocity = u

Let the length of the train be L


L = [ (3u² - u²) ] / 2a

L = ( 9u² - u²)/2a

L = 8u²/2a

L = 4u²/a ----- (i)

Now,

Midpoint will be at the distance L/2

Using equation (i)

4u² ÷ a/2 = (v² - u²)/2a

2u²/a = (v² - u²)/2a

2u² = (v² - u²)/2

4u² = v² - u²

4u² + u² = v²

5u² = v²


v = u√5 or √5 u

Thus, the velocity with which the middle point of the train passes the same point is √5 u.
Option (3) is correct.

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Hope it helps...!!!