Question

A body moving with a constant acceleration has velocities u and v when passing through points A and B in its path . The velocity of the body midway between A and B is
a.(u+v)/2
b.under root (u²+v²)/2
c.(u+at)/2
d. none of these

I know the option it's 'b' .
But I need a solution .

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the Velocity at A be "u".
• Let the velocity at B be "v".
• Let the velocity at mid - point be "K".

$\huge{\bold{\underline{Explanation:-}}}$

Applying third kinematic equation,

$\large{\bold{v^2 - u^2 = 2as}}$

Taking displacement,

$\large{\boxed{S = \dfrac{v^2 - u^2}{2a}}}$

$\large{ S = \dfrac{v^2 - u^2}{2a} \: ----(1)}$

Now, For mid - point the Displacement will be h = S/2.

Applying third kinematic equation,

$\large{\bold{v^2 - u^2 = 2as}}$

Substituting the values,

$\large{K^2 - u^2 = 2 \times a \times h}$

h = S/2.

$\large{K^2 - u^2 = 2 \times a \times \dfrac{S}{2}}$

$\large{K^2 - u^2 = \cancel{2} \times a \times \dfrac{S}{ \cancel{2}}}$

$\large{K^2 - u^2 = aS}$

Substitute the value of "S" from equation(1) in the above equation.

$\large{K^2 - u^2 = a \times \dfrac{v^2 - u^2}{2a}}$

$\large{K^2 - u^2 = \cancel{a} \times \dfrac{v^2 - u^2}{2 \cancel{a}}}$

$\large{K^2 - u^2 = \dfrac{v^2 - u^2}{2}}$

$\large{K^2 = u^2 + \dfrac{v^2 - u^2}{2}}$

Taking L.C.M

$\large{K^2 = \dfrac{2u^2 + v^2 - u^2}{2}}$

$\large{K^2 = \dfrac{v^2 + u^2}{2}}$

$\huge{\boxed{\boxed{K = \sqrt{ \dfrac{v^2 + u^2}{2}}}}}$

Hence derived.

So,the option (B) is correct.