Question

Q1.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

Step-by-step explanation:

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:

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