Question

The distance traveled by a body in $I{V}^{th}$ second is twice the distance traveled in $I{I}^{nd}$ second. If the acceleration of the body is $3m/s^2$, then its initial velocity is

• A. $\frac{3}{2}m/s$
• B. $\frac{5}{2}m/s$
• C. $\frac{7}{2}m/s$
• D. $\frac{9}{2}m/s$

Answer 1

The correct option is A $\frac{3}{2}m/s$

$s_4=2s_2(4u+\frac{a}{2}{4}^2)-(3u+\frac{a}{2}{3}^2)=2\left\{\right(2u+\frac{a}{2}{2}^2)-(u+\frac{a}{2}\}\Rightarrow 10.5-9=u\Rightarrow u=\frac{3}{2}m/s$