Question

Q1. A body moving with a uniform acceleration crosses a distance of 65m in 5th second and 105m in 9th second. How far will it go in 20th is seconds and what is the distance covered in 20th second?

Answer 1

Let u and a be the initial velocity and uniform acceleration of the body.

$$\begin{gathered} D_n=u+\frac{a}{2}\left(2n-1\right) \\ \therefore D_5=u+\frac{a}{2}\left(2\times 5-1\right) \\ \Rightarrow 65=u+\frac{9a}{2}.....\left(1\right) \\ \mathrm{Similarly} \\ 105=u+\frac{17a}{2}.....\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right)\mathrm{we}\mathrm{get} \\ 80=8a \\ \Rightarrow a=10\mathrm{m}{\mathrm{s}}^{-2} \\ \mathrm{Substituting}\mathrm{this}\mathrm{value}\mathrm{of}\mathit{}a\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get} \\ 65=u+45 \\ \Rightarrow u=20\mathrm{m}{\mathrm{s}}^{-1} \\ \mathrm{Distance}\mathrm{at}20\mathrm{second}\mathrm{will}\mathrm{be} \\ {D}_{20}=u+\frac{a}{2}\left(2\times 20-1\right) \\ \Rightarrow D_{20}=20+\frac{390}{2} \\ \Rightarrow D_{20}=20+195=215\mathrm{m} \\ \mathrm{Distance}\mathrm{covered}\mathrm{in}20\mathrm{second}\mathrm{will}\mathrm{be} \\ S=ut+\frac{1}{2}a{t}^2 \\ \mathit{\Rightarrow }S=20\times 20+\frac{1}{2}\times 10\times {20}^2 \\ \Rightarrow S=400+2000=2400\mathrm{m} \end{gathered}$$