Motion in a Straight Line - Result Question 18

18. The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} ms^{-2}$, in the third second is:

[2008]

(a) $6 m$

(b) $4 m$

(c) $\frac{10}{3} m$

(d) $\frac{19}{3} m$

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Answer:

Correct Answer: 18. (c)

Solution:

  1. (c) Distance travelled in the nth second is given by $S_n=u+\frac{a}{2}(2 n-1)$

Given $u=0, a=\frac{4}{3} ms^{-2}, n=3$

$\therefore S_n=0+\frac{4}{3 \times 2}(2 \times 3-1)=\frac{4}{6} \times 5=\frac{10}{3} m$



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