A particle move in straight line with initial velocity'u' and acceleration'a'. Derive the equation for it's displacement in nth second
Share
A particle move in straight line with initial velocity'u' and acceleration'a'. Derive the equation for it's displacement in nth second
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
Given:
A particle move in straight line with initial velocity 'u' and acceleration 'a'.
To find:
Equation for it's displacement in nth second ?
Calculation:
The displacement in the "n"th second can be calculated by subtracting the position at "(n-1)" time from "(n)" time.
[tex] \therefore \: s_{n} = ut + \dfrac{1}{2} a {t}^{2} [/tex]
[tex] \implies \: s_{n} = u(n)+ \dfrac{1}{2} a {(n)}^{2} [/tex]
[tex] \implies \: s_{n} = un+ \dfrac{1}{2} a {n}^{2} \: \: \: \: .......(1)[/tex]
Now, at t = (n-1) sec:
[tex] \therefore \: s_{(n - 1)} = ut + \dfrac{1}{2} a {t}^{2} [/tex]
[tex] \implies \: s_{(n - 1)} = u(n - 1) + \dfrac{1}{2} a {(n - 1)}^{2} [/tex]
[tex] \implies \: s_{(n - 1)} = u(n - 1) + \dfrac{1}{2} a ( {n}^{2} - 2n + 1)[/tex]
[tex] \implies \: s_{(n - 1)} = un - u + \dfrac{1}{2} a ( {n}^{2} - 2n + 1) \: \: \: \: ......(2)[/tex]
Subtracting the equations:
[tex] \therefore \: s_{ {n}^{th} } = s_{(n )} - s_{(n - 1)}[/tex]
[tex] \implies \: s_{ {n}^{th} } = \bigg \{un + \dfrac{1}{2} a {n}^{2} \bigg \} - \bigg \{un - u + \dfrac{1}{2} a( {n}^{2} - 2n + 1) \bigg \} [/tex]
[tex] \implies \: s_{ {n}^{th} } = u + \dfrac{1}{2} a(2n - 1)[/tex]
So, final expression:
[tex] \boxed{ \bf\: s_{ {n}^{th} } = u + \dfrac{1}{2} a(2n - 1)}[/tex]