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

Answer: (B)

18 m

Answer 2

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, $t=2,\,\,s=3\times \frac{{{2}^{3}}}{3}+\frac{5\times {{2}^{2}}}{2}$ $=3\times \frac{8}{3}+\frac{5\times 4}{2}$ $=8+10=18\,\,m$