Question

A particle starting from rest begins to move along a straight line with acceleration a. Its complete path x is divided into n equal parts. At the end of each part, its acceleration increases by a/n. Prove that the velocity of the particle at the end of the path is √[ax(3n – 1) /n]​

Answer 1

Answer:

A particle starting from rest begins to move along a straight line with acceleration a. Its complete path x is divided into n equal parts. At the end of each part, its acceleration increases by a/n. Prove that the velocity of the particle at the end of the path is √[ax(3n – 1) /n]sh the following the best thing I am looking at it was the last one or the not the case I am not too far away in a different story than a good day and age group for a new thread for a you have the opportunity for you and me to come to our website is the first one or not I was just thinking about doing the first two weeks to be honest and it has a great weekend so much we have had no problem and a great weekend so we will get the idea that is not working with in your company that you would be want a bit to get to see what they want a bit

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If

Initial Velocity = u

Final Velocity = v

Acceleration = a

Distance covered = s

Then

${v}^{2} = {u}^{2} + 2as$

$\implies \: {v}^{2} - {u}^{2} = 2as \: \: ...(1)$

EVALUATION

Let starting point = A

Finishing point = B

The distance between A & B is divided into n equal parts by the points

$P_1, P_2, P_3,...... P_{n-1}$

So that

$\displaystyle \: P_1-A= P_2-P_1=.....=B-P_{n-1} = \frac{x}{n}$

Also let the velocity at the points

$P_1, P_2, P_3,...... P_{n-1} \: be \:$

$u,v_1, v_2, v_3 \: ,...... v_{n-1},v \: \: respectively$

Now

$\sf{\underline{ \: When \: particle \: arrives \: at \: \: P_1}} :$

$Initial \: Velocity = u = 0$

$Final \: Velocity = v_1$

$Acceleration = a$

$\displaystyle \: Distance \: covered = \: \frac{x}{n}$

So using (1)

$\displaystyle \: {v_1}^{2} - {u}^{2} = 2a \times \frac{x}{n} \: \: ...(2)$

$\sf{\underline{ \: When \: particle \: arrives \: at \: \: P_2}} :$

$Initial \: Velocity = v_1$

$Final \: Velocity = v_2$

$\displaystyle \: Acceleration = a + \frac{a}{n}$

$\displaystyle \: Distance \: covered = \: \frac{x}{n}$

Using (1)

$\displaystyle \: {v_2}^{2} - {v_1}^{2} = 2(a + \frac{a}{n} )\times \frac{x}{n} \: \: ...(3)$

$\sf{\underline{ \: When \: particle \: arrives \: at \: \: P_3}} :$

$Initial \: Velocity = v_2$

$Final \: Velocity = v_3$

$\displaystyle \: Acceleration = a + \frac{2a}{n}$

$\displaystyle \: Distance \: covered = \: \frac{x}{n}$

Using (1)

$\displaystyle \: {v_3}^{2} - {v_2}^{2} = 2(a + \frac{2a}{n} )\times \frac{x}{n} \: \: ...(4)$

.

.

.

.

.

$\sf{\underline{ \: When \: particle \: arrives \: at \: B \:}} :$

$Initial \: Velocity = v_{n - 1}$

$Final \: Velocity = v$

$\displaystyle \: Acceleration = a + \frac{(n - 1)a}{n}$

$\displaystyle \: Distance \: covered = \: \frac{x}{n}$

Using (1)

$\displaystyle \: {v}^{2} - {v_{n - 1}}^{2} = 2(a + \frac{(n - 1)a}{n} )\times \frac{x}{n} \: \: ...(n)$

Adding Equation (2) + (3) +..... +(n) we get

${v}^{2} - {u}^{2} = \displaystyle 2a\times \frac{x}{n} + 2(a + \frac{a}{n} )\times \frac{x}{n} + 2(a + \frac{2a}{n} )\times \frac{x}{n} + .. +2(a + \frac{(n - 1)a}{n} )\times \frac{x}{n}$

$\implies \: {v}^{2} = \displaystyle \frac{2x}{n} \bigg( a + (a + \frac{a}{n} )+ (a + \frac{2a}{n} ) + .. +(a + \frac{(n - 1)a}{n} ) \bigg)$

$\implies \: {v}^{2} = \displaystyle \frac{2x}{n} \bigg( \frac{n}{2}(2a + (n - 1) \times \frac{a}{n} \bigg)$

It is a arithmetic progression

So

$\implies \: {v}^{2} = \displaystyle xa(2 + \frac{n - 1}{n} )$

$\implies \: {v}^{2} = \displaystyle xa \times \frac{3n - 1}{n} \times$

$\implies \: {v} = \displaystyle \sqrt{ \frac{xa(3n - 1)}{n} }$

So the final velocity is

$\: \displaystyle \sqrt{ \frac{xa(3n - 1)}{n} }$

Hence proved