Q124 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Physics
The acceleration of a particle is increasing linearly with time $t$ as $6 t$. The particle starts from the origin with an initial velocity $10 \mathrm{~m} / \mathrm{s}$. The distance travelled by the particle after $2 \mathrm{~s}$ will be
- A18 m
- Correct answerB44 m
- C22 m
- D26 m
Solution
Acceleration of the particle, $a=6 t$
Initial velocity, $u=10 \mathrm{~m} / \mathrm{s}$
Let the distance travelled after $t=2 \mathrm{~s}$ is $s$.
$$
\begin{aligned}
\therefore \quad s & =\left(u \cdot t+\frac{1}{2} a t^2\right) \\
& =\left(10 t+\frac{1}{2} \times 6 t \times t^2\right) \\
& =10 \times 2+\frac{6}{2} \times 2^3 \\
s & =20+24=44 \mathrm{~m}
\end{aligned}
$$
$\left({ }^*\right)$ None option is correct.
Hence, the correct answer is option: 'B'.