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 $3rd$ is

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

Answer 1

The correct option is B

$40\%$

Step1: Given data and assumptions

A particle moves with constant acceleration = $a$

Initial velocity, $u=0$

Step2: Finding the percentage increase in displacement during the $4th$ second compared to that in the $3rd$ second.

We know that.

The displacement of the particle in $nth$ second is

$S_n=u+\frac{1}{2}a(2n-1)$

Where $S_n$ $nth$ term displacement, $u$ is initial velocity, $a$ is acceleration.

Now,

The displacement in the $3rd$ second.

$S_3=\frac{1}{2}a(2\times 3-1)=\frac{5a}{2}$$\left[\because u=0\right]$ …..$\left(i\right)$

And the displacement in the $4th$second.

$S_4=\frac{1}{2}a(2\times 4-1)=\frac{7a}{2}$ …..$\left(ii\right)$

Therefore,

The percentage increase in the displacement $4th$ second compared to that in the $3rd$ second.

$\frac{∆S}{S}\%=\frac{S_4-S_3}{S_3}\times 100$

$\Rightarrow$$\frac{∆S}{S}\%=\frac{{\displaystyle \frac{7a}{2}}-{\displaystyle \frac{5a}{2}}}{\frac{5a}{2}}$ $\left[\because From\left(i\right)and\left(ii\right)\right]$

$\Rightarrow$$\frac{∆S}{S}\%=40\%$

Hence, option B is correct.