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The distance travelled by a particle starting from rest and moving with an acceleration $\dfrac{4}{3}$ m s$^{-2}$, in the third second is -
A
$\dfrac{10}{3}$ m
B
$\dfrac{19}{3}$ m
C
6 m
D
4 m
Detailed Solution
Distance covered in the $n$th second: $S_n = u + \dfrac{a}{2}(2n - 1)$
Here $u = 0$, $a = \dfrac{4}{3}$ m s$^{-2}$ and $n = 3$.
$S_3 = 0 + \dfrac{1}{2} \times \dfrac{4}{3} \times (2 \times 3 - 1)$
$S_3 = \dfrac{1}{2} \times \dfrac{4}{3} \times 5$
$S_3 = \dfrac{10}{3}$ m
Check: distance in 3 s $= \dfrac{1}{2}\cdot\dfrac{4}{3}\cdot 9 = 6$ m and in 2 s $= \dfrac{1}{2}\cdot\dfrac{4}{3}\cdot 4 = \dfrac{8}{3}$ m; the difference is $6 - \dfrac{8}{3} = \dfrac{10}{3}$ m.
Here $u = 0$, $a = \dfrac{4}{3}$ m s$^{-2}$ and $n = 3$.
$S_3 = 0 + \dfrac{1}{2} \times \dfrac{4}{3} \times (2 \times 3 - 1)$
$S_3 = \dfrac{1}{2} \times \dfrac{4}{3} \times 5$
$S_3 = \dfrac{10}{3}$ m
Check: distance in 3 s $= \dfrac{1}{2}\cdot\dfrac{4}{3}\cdot 9 = 6$ m and in 2 s $= \dfrac{1}{2}\cdot\dfrac{4}{3}\cdot 4 = \dfrac{8}{3}$ m; the difference is $6 - \dfrac{8}{3} = \dfrac{10}{3}$ m.