Question

A particle starts with an initial velocity and passes successively over the two halves of a given distance with accelerations $a_1$ and $a_2$ respectively. Show that the final velocity is the same as if the whole distance is covered with a uniform acceleration $\frac{(a_1+a_2)}{2}.$

Answer 1

Let the initial velocity in both cases be $u$

First case :

For first half,

Apply 3rd equation of motion-

$v^2=u^2+2aS$

$v$ : Final velocity of the particle

$u$ : Initial velocity of the particle

$a$ : Acceleration of the particle

$S$ : Displacement of the oarticle

$v_1^2-u^2=2a_{1}x$ ..........(1)

For another half,

$v_2^2-v_1^2=2a_{2}x$ ...............(2)

Adding (1) and (2) we get

$v_2^2-u^2=2(a_1+a_2)x$

$⟹v_2=\sqrt{u^2+2(a_1+a_2)x}$ ............(a)

Second case :

$a_t=\frac{a_1+a_2}{2}$

$S=2x$

$\therefore V^2-u^2=2(\frac{a_1+a_2}{2})2x$

$⟹$ $V=\sqrt{u^2+2(a_1+a_2)x}$ ..................(b)

Thus from (a) and (b), the final velocity is same in both cases.