Question

A particle moves along a straight line AB with constant acceleration . Its. velocities are u and v at A and B respectively. Show that its velocity at the mid - point of AB is √ u^ 2 + v ^ 2 / 2 .

Answer 1

Heya.........!!!!

______________________________


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 ( velocity at midpoint of AB ) =>

$\sqrt{ \frac{v {}^{2} + \: u {}^{2} }{2} }$


==================================

Hope It Helps You ☺

Answer 2

Answer:

Hope you have received your answer...