Q154 / 170 · Manipal (MET) 2021 · Physics
A point initially at rest moves along $x$-axis. Its acceleration varies with time as $a=(6 t+5)$ in $\mathrm{m} / \mathrm{s}^2$. If it starts from origin, the distance covered in 2 s is :
- A20 m
- Correct answerB18 m
- C16 m
- D25 m
Solution
Given, $a=\frac{d v}{d t}=6 t+5$ Integrating it, $\int_0^v d v=\int_0^t(6 t+5) d t$
$v=\frac{6 t^2}{2}+5 t$
As $\quad v=\frac{d s}{d t}$, so $d s=\left(\frac{6 t^2}{2}+5 t\right) d t$ Integrating it, $\int_0^s d s=\int_0^t\left(\frac{6 t^2}{2}+5 t\right) d t$
$s=\frac{3 t^3}{3}+\frac{5 t^2}{2}$
When $\begin{aligned} t & =2 \mathrm{~s}, s=3 \frac{(2)^3}{3}+\frac{5(2)^2}{2} \\ & =8+10=18 \mathrm{~m}\end{aligned}$
Hence, the correct answer is option: 'B'.