Question

Solve this:

Q1. The displacement x of a particle moving along x-axis at time t is given by x² = 2t² + 6t. The velocity at any time t is


$$\begin{gathered} \left(1\right)\frac{4t+6}{x}\left(2\right)\frac{2t+x}{x} \\ \left(3\right)4t+6\left(4\right)\frac{2t+3}{x} \end{gathered}$$

Answer 1

Dear Student,

$$\begin{gathered} {\mathrm{x}}^2=2{\mathrm{t}}^2+6\mathrm{t} \\ \mathrm{x}={\left(2{\mathrm{t}}^2+6\mathrm{t}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2{\mathrm{t}}^2+6\mathrm{t}\right) \\ \mathrm{v}=\frac{1}{2}{\left(2{\mathrm{t}}^2+6\mathrm{t}\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[2\left(2\mathrm{t}\right)\left(2\right)+6\left(1\right)\right] \\ \mathrm{v}=\frac{1}{2}\frac{1}{\sqrt{2{\mathrm{t}}^2+6\mathrm{t}}}\times \left[8\mathrm{t}+6\right] \\ \mathrm{v}=\frac{4\mathrm{t}+3}{\mathrm{x}} \end{gathered}$$