Question

Hieee :]
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$\huge\dag \: \underline\mathfrak{Question :-}$
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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. .?
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Answer with explanation! ​

Answer 1

$\rm \large \red {\boxed{Helloww!}}$

$\rm \large {\underbrace{\underline{Elucidation:-}}}$

$\sf \blue {\underline{\underline{Given:}}}$

->The velocity of train with which it passes the electric pole=u

->The velocity of last compartment=v

->Acceleration=uniform

$\sf \pink {\underline{\underline{To\: Determine:}}}$

->The velocity of the middle part of the train that passes through the same pole=?

$\sf \purple {\underline{\underline{Kinematic\: Equation\: 2\: b\: used:}}}$

$\sf {\boxed{\underline{v^{2}=u^{2}+2as}}}$

->Here the distance 's' is considered as the length of train ; and let it be 'x'

$\large \sf \mapsto {v^{2}=u^{2}+2ax}$

$\large \sf \mapsto {v^{2}-u^{2}=2ax}$

$\large \sf \mapsto {a=\frac{v^{2}-u^{2}}{2x}}$------Eq(1)

->If this is the equation obtained when the train passes the pole,then the velocity of middle part that passes the same pole will be ,

->Again by using the same kinematic equation ,

$\sf {v^{2}=u^{2}+2as}$

-> If the length of the train is x, the length of middle part of the train is $\sf \large{\frac{x}{2}}$

$\sf {v^{2}=u^{2}+2a(\frac{x}{2})}$--------Eq(2)

$\sf \orange {\underline{\underline{Substitution\:nd\: Solving:}}}$

->Substituting Equation(1) in Equation(2),

$\large \sf \mapsto {v^{2}=u^{2}+2(\frac{v^{2}-u^{2}}{2x})(\frac{x}{2})}$

$\large \sf \mapsto {v^{2}=u^{2}+\cancel{2}(\frac{v^{2}-u^{2}}{2\cancel{x}})(\frac{\cancel{x}}{\cancel{2}})}$

$\large\sf \mapsto {v^{2}=u^{2}+\frac{v^{2}-u^{2}}{2}}$

$\large \sf \mapsto {v^{2}=\frac{2u^{2}+v^{2}-u^{2}}{2}}$

$\large \sf \mapsto {v^{2}=\frac{v^{2}+u^{2}}{2}}$

$\large \sf \mapsto {v=\sqrt{\frac{v^{2}+u^{2}}{2}}}$

$\sf \green {\underline{\underline{Henceforth:}}}$

->The velocity of the middle part of the train that passes through the same pole turns out to be ,

$\large \sf \implies {\boxed{v=\sqrt{\frac{v^{2}+u^{2}}{2}}}}$

Answer 2

Answer:

(1) length of train :- l

When last compartment passes through pole distance covered by first compartment is l

∴v

2

=u

2

+2al

∴a=

2l

v

2

−u

2

Now, when middle post passes, distance covered by first compartment is

2

l

∴v

f

2

=u

2

+

2

Ral

=u

2

+(

2l

v

2

−u

2

)l=

2

v

2

+u

2

∴v

f

=

2

v

2

+u

2