Physics, asked by abhi0013, 1 year ago

please answer the question with proper explanation

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Answered by rohit710
21
Heya......!!!

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Let the total distance be ( s ) .

let the points be A , B and midpoint = C

• velocity at point A = v1
• velocity at point B = v2

Let Velocity at point C ( midpoint ) = x .

Between A and B :-

=> v2^2 = v1^2 + 2as
=> v2^2 - v1^2 = 2as

=> as = (v2^2 - v1^2) / 2

Between A and C :-

=> x^2 = v1^2 + 2as/2 ( s = s/2 as midpoint ) .
=> x^2 = v1^2 + as

putting value of ' as ' in this equation.

=> x^2 = v1^2 + ( v2^2 - v1^2 ) / 2
=> x^2 = ( v1^2 + v2^2 ) / 2

 \\ x \: = \: \sqrt{ \frac{v_{1 {}} {}^{2} + v_ {2}^{2} } {2} }

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Hope It Helps You ☺

PrincessNumera: nice newton ;p
rohit710: :)Thanks ;-)
Anonymous: awesome bro
rohit710: Thanku Sis :D
Answered by BrainlyFlash156
0

\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ĄNsWeR࿐}}}

Let the total distance be ( s ) .

let the points be A , B and midpoint = C

• velocity at point A = v1

• velocity at point B = v2

Let Velocity at point C ( midpoint ) = x .

Between A and B :-

=> v2^2 = v1^2 + 2as

=> v2^2 - v1^2 = 2as

=> as = (v2^2 - v1^2) / 2

Between A and C :-

=> x^2 = v1^2 + 2as/2 ( s = s/2 as midpoint ) .

=> x^2 = v1^2 + as

putting value of ' as ' in this equation.

=> x^2 = v1^2 + ( v2^2 - v1^2 ) / 2

=> x^2 = ( v1^2 + v2^2 ) / 2

 \\ x \: = \: \sqrt{ \frac{v_{1 {}} {}^{2} + v_ {2}^{2} } {2} }

HOPE SO IT WILL HELP......

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