Question

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:

• A. 33%
• B. 77%
• C. 66%
• D. 40%

Answer 1

Answer: (D)

The correct option is B 40%
The displacement of the particle in $nth$ second $S_n=u+\frac{1}{2}a(2n-1)$

where all the symbols have their usual meaning.

Given : $u=0$

$⟹S_n=\frac{1}{2}a(2n-1)$

Thus displacement in the $3rd$ second $S_3=\frac{1}{2}a(2\times 3-1)=\frac{5a}{2}$

Displacement in the $4th$ second $S_4=\frac{1}{2}a(2\times 4-1)=\frac{7a}{2}$

Percentage increase in the displacement $\frac{\Delta S}{S_3}\times 100=\frac{S_4-S_3}{S_3}\times 100$

$\frac{\Delta S}{S_3}\times 100=\frac{\frac{7a}{2}-\frac{5a}{2}}{\frac{5a}{2}}\times 100=40$

Thus percentage increase in the displacement is $40$ %