Question

acceleration of a particle is a for a time t. It is followed immediately by retardation of a for time t/2 . consider this as one cycle. initial velocity of the particle was zero. the displacement of the particle after n such cycles in successions is

Answer 1

I have assumed alpha as acceleration and deceleration.
Refer to photos for answer
very much sorry for the resolution !

Answer 2

Answer:

The displacement of the particle is given as n(3n + 4)at²/8.

Explanation:

It is given that the particle has an acceleration 'a' for time t and retardation '-a' for time t/2.

So, the velocity at the end of (r-1)th cycle is given as

v1 =(r-1) [u + at]

Since initial velocity is zero, i.e., u = 0. Therefore,

v1 = (r-1)[at - at/2]

v1 = (r-1)(at/2)

Velocity at the end of (accelerating half) rth cycle is

v2 = u + at

v2 = (r-1)(at/2) + at

v2 = (r+1)(at/2)

Now, the total displacement in rth cycle is

$S=S_{1half} + S_{2half}$

Displacement is given as $S = ut + \frac{1}{2}at^2$

$S = (r-1)\frac{at}{2}t + \frac{1}{2}at^2 + (r+1)\frac{at}{2}\frac{t}{2}- \frac{1}{2}a(\frac{t}{2}) ^2$

$S = \frac{rat^2}{2} - \frac{at^2}{2} + \frac{at^2}{2} +\frac{rat^2}{4}+\frac{at^2}{4} - \frac{at^2}{8}$

$S = \frac{3rat^2}{4}+ \frac{at^2}{8}$

$S = \frac{at^2}{8} (6r+1)$

Therefore, displacement in the nth cycle is

Sₙ = Σ$\frac{at^2}{8} (6r+1)$

Sₙ = n(3n + 4)at²/8

Therefore, the displacement of the particle after n cycles in succession is n(3n + 4)at²/8.

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