Question

A body moving with uniform acceleration describes 12m in 3rd second and 20m in 5th second. Find the velocity after 10second?

Answer 1

$$\begin{gathered} \mathrm{Distance}\mathrm{travelled}\mathrm{in}\mathrm{nth}\mathrm{second}: \\ {\mathrm{S}}_{\mathrm{n}}=\mathrm{u}+\mathrm{a}\left(\mathrm{n}-\frac{1}{2}\right) \\ \mathrm{Case}1:3\mathrm{rd}\mathrm{second} \\ 12=\mathrm{u}+\mathrm{a}\left(3-\frac{1}{2}\right) \\ 12=\mathrm{u}+2.5\mathrm{a}......\left(1\right) \\ \mathrm{Case}2:5\mathrm{th}\mathrm{second} \\ 20=\mathrm{u}+\mathrm{a}\left(5-\frac{1}{2}\right) \\ 20=\mathrm{u}+5.5\mathrm{a}......\left(2\right) \\ \left(2\right)-\left(1\right): \\ 8=3\mathrm{a} \\ \mathrm{a}=\frac{8}{3}\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Then},\mathrm{u}=12-\left(2.5\times \frac{8}{3}\right) \\ =\frac{36-20}{3} \\ =\frac{16}{3}=5.33\mathrm{m}/\mathrm{s} \\ \mathrm{Case}3:\mathrm{Velocity}\mathrm{after}10\mathrm{seconds} \\ \mathrm{v}=\mathrm{u}+\mathrm{at} \\ \mathrm{v}=\frac{16}{3}+\frac{8}{3}\times 10 \\ \mathrm{v}=\frac{16+80}{3}=32\mathrm{m}/\mathrm{s} \end{gathered}$$