Question

A car starts from rest and moves with constant accleration . The ratio of distance covered by the car in nth sec to that covered in n second is

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{initial}\mathrm{velocity},\mathrm{u}=0 \\ \mathrm{Distance}\mathrm{covered}\mathrm{in}{\mathrm{n}}^{\mathrm{th}}\mathrm{second}, \\ {\mathrm{S}}_{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}\left(2\mathrm{n}-1\right) \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}}{2}\left(2\mathrm{n}-1\right) \\ \mathrm{Distance}\mathrm{covered}\mathrm{in}\mathrm{n}\mathrm{seconds}, \\ \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{S}=0\times \mathrm{n}+\frac{\mathrm{a}}{2}{\mathrm{n}}^2 \\ \mathrm{S}=\frac{\mathrm{a}}{2}{\mathrm{n}}^2 \\ \frac{{\mathrm{S}}_{\mathrm{n}}}{\mathrm{S}}=\frac{\frac{\mathrm{a}}{2}\left(2\mathrm{n}-1\right)}{\frac{\mathrm{a}}{2}{\mathrm{n}}^2} \\ \frac{{\mathrm{S}}_{\mathrm{n}}}{\mathrm{S}}=\frac{\left(2\mathrm{n}-1\right)}{{\mathrm{n}}^2} \end{gathered}$$