Question

An object is moving along a straight line with initial velocity u and constant acceleration a. What is the displacement of the particle in the nth second?​

Answer 1

Given:

An object is moving along a straight line with initial velocity u and constant acceleration a.

To find:

Displacement in nth second?

Calculation:

• Displacement in nth second can be calculated by subtracting the position of the object at (n-1)th second from nth second.

At t = n - 1

$\rm \: x_{n - 1} = u(n - 1)+ \dfrac{1}{2} a {(n - 1)}^{2}$

At t = n

$\rm \: x_{n} = un + \dfrac{1}{2} a {(n)}^{2}$

Subtracting the Equation

$\rm \: x_{ {n}^{th} } = x_{n} - x_{(n-1)}$

$\rm \implies \: x_{ {n}^{th} } =un + \dfrac{1}{2} a {n}^{2} - \bigg\{u(n-1) + \dfrac{1}{2}a(n-1)^{2}\bigg\}$

$\rm \implies \: x_{ {n}^{th} } = u + \dfrac{1}{2} a {n}^{2} -\dfrac{1}{2} a {(n -1)}^{2}$

$\rm \implies \: x_{ {n}^{th} } = u + \dfrac{1}{2} a \bigg \{{ {n}^{2}- (n -1)}^{2} \bigg \}$

$\rm \implies \: x_{ {n}^{th} } = u + \dfrac{1}{2} a \bigg \{ {n}^{2} - {n}^{2} +2n-1 \bigg \}$

$\rm \implies \: x_{ {n}^{th} } = u + \dfrac{1}{2} a \bigg \{ 2n -1 \bigg \}$

So, final answer is:

$\boxed{ \bf\: x_{ {n}^{th} } = u + \dfrac{1}{2} a \bigg \{ 2n -1 \bigg \}}$

Answer 2

Given : An object is moving along a straight line with initial velocity u and constant acceleration a.

To Find :  displacement of the particle in the nth second?​

Solution :

Sₙ  - Sₙ₋₁

S = ut + (1/2)at²

= un + (1/2)a(n)²  -  (u(n-1) + (1/2)a(n-1)²)

= u +  (1/2)a ( n² - (n - 1)²)

= u + (1/2)a (2n - 1)(1)

= u + (a/2)(2n - 1)  

displacement of the particle in the nth second = u +  (a/2) (2n - 1)  

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