Q105 / 200 · NEET 2008 (Screening) · Physics
The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} \mathrm{~ms}^{-2}$, in the third second is
- A$6 \mathrm{~m}$
- B$4 \mathrm{~m}$
- Correct answerC$\frac{10}{3} \mathrm{~m}$
- D$\frac{19}{3} \mathrm{~m}$
Solution
Distance travelled by the particle in $n$th second is
$S_{n t h}=u+\frac{1}{2} a(2 n-1)$
where $u$ is initial speed and $a$ is acceleration of the particle.
Here, $n=3, u=0, a=\frac{4}{3} \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned}
\therefore S_{3 \text { rd }} & =0+\frac{1}{2} \times \frac{4}{3} \times(2 \times 3-1) \\
& =\frac{4}{6} \times 5 \\
& =\frac{10}{3} \mathrm{~m}
\end{aligned}$
Alternative : Distance travelled in the 3rd second $=$ distance travelled in $3 \mathrm{~s}$ - distance travelled in $2 \mathrm{~s}$
As, $u=0$,
$S_{(3 \mathrm{rd} s)}=\frac{1}{2} a \cdot 3^2-\frac{1}{2} a \cdot 2^2=\frac{1}{2} \cdot a \cdot 5$
Given $a=\frac{4}{3} \mathrm{~ms}^{-2}$
$\therefore S_{(3 \mathrm{rd} s)}=\frac{1}{2} \times \frac{4}{3} \times 5=\frac{10}{3} \mathrm{~m}$
Hence, the correct answer is option: 'C'.