Question

A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the $4^{th}$ second compared to that in the $3^{rd}$ second is

• A. $33\%$
• B. $40\%$
• C. $66\%$
• D. $77\%$

Answer 1

Answer: (B)

$40\%$

Answer 2

We know that
$S_{n t h}=u+\frac{1}{2} a(2 n-1)$
$S_{3 \text { rd }}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a ($ for $n=3 s)$
$S_{4 t h}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a ($ for $n=4 s )$
So, the percentage increase
$=\frac{S_{4 h }-S_{3 rd }}{S_{3 rd }} \times 100$
$=\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100$
$=\frac{\frac{2 a}{2}}{\frac{5}{2} a} \times 100$
$=2 \times 20=40 \%$