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A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second is
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Verified Answer
The correct answer is:
$40 \%$
We know that
$S=u+\frac{1}{2} a(2 n-1)$
$S_{3rd \text { sec}}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a$ (for $n=3$ s)
$S_{4th \text { sec}}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a \quad($ for $n=4 \mathrm{s})$
So, the percentage increase
$=\frac{S_{4 t h}-S_{3 \mathrm{rd}}}{S_{3 \mathrm{rd}}} \times 100$
$\begin{array}{l} =\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100 \\ =\frac{\frac{2 a}{2}}{\frac{5}{2}} \times 100=2 \times 20=40 \% \end{array}$
$S=u+\frac{1}{2} a(2 n-1)$
$S_{3rd \text { sec}}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a$ (for $n=3$ s)
$S_{4th \text { sec}}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a \quad($ for $n=4 \mathrm{s})$
So, the percentage increase
$=\frac{S_{4 t h}-S_{3 \mathrm{rd}}}{S_{3 \mathrm{rd}}} \times 100$
$\begin{array}{l} =\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100 \\ =\frac{\frac{2 a}{2}}{\frac{5}{2}} \times 100=2 \times 20=40 \% \end{array}$
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