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)

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