Question

A point initially at rest moves along $x$-axis. Its acceleration varies with time as $ a=(6t+5)\,m/s^{2}$ . If it starts from origin, the distance covered in $2\, s$ is

• A. 20 m
• B. 18 m
• C. 16 m
• D. 25 m

Answer 1

The Correct Option is B

Given, $ a=\frac{dv}{dt}=6t+5 $ or $ dv=(6t+5)dt $ Integrating, we get $ \int\limits_{0}^{v}{dv}=\int\limits_{0}^{1}{(6t+5)}dt $ or $ v=\left( \frac{6{{t}^{2}}}{2}+5t \right) $ Again $ v=\frac{ds}{dt} $ $ \therefore $ $ ds=\left( \frac{6{{t}^{2}}}{2}+5t \right)dt $ Integrating again, we get $ \int\limits_{0}^{s}{ds}=\int\limits_{0}^{1}{\left( \frac{6{{t}^{2}}}{2}+5t \right)dt} $ $ \therefore s=\frac{3{{t}^{3}}}{3}+\frac{5{{t}^{2}}}{2} $ When, $ t=2,\,\,s=3\times \frac{{{2}^{3}}}{3}+\frac{5\times {{2}^{2}}}{2} $ $ =3\times \frac{8}{3}+\frac{5\times 4}{2} $ $ =8+10=18\,\,m $