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:
- (c) Distance travelled in the nth second is given by $S_n=u+\frac{a}{2}(2 n-1)$
Given $u=0, a=\frac{4}{3} ms^{-2}, n=3$
$\therefore S_n=0+\frac{4}{3 \times 2}(2 \times 3-1)=\frac{4}{6} \times 5=\frac{10}{3} m$
