Question

A body covers a distance of $4\, m$ in $3^{rd}$ second and $12 \,m$ in $5^{th}$ second. If the motion is uniformly accelerated, how far will it travel in the next $3$ seconds ?

• A. $10\,m$
• B. $30\,m$
• C. $40\,m$
• D. $60\,m$

Answer 1

Answer: (D)

$S_{3}=u+\frac{a}{2}\left(2\times3-1\right)=4$ or $u+\frac{5}{2} a=4$
$S_{5}=u+\frac{a}{2}\left(2\times5-1\right)=12$ or $u+\frac{9}{2}a=12$
On solving, $u=-6\,m\,s^{-1}, a=4\,m\,s^{-2}$
Distance travelled in next $3$ seconds $=S_{8}-S_{5}$
$=\left[-6\times8+\frac{1}{2}\times4\times\left(8\right)^{2}\right]-\left[-6\times5+\frac{1}{2}\times4\times\left(5\right)^{2}\right]$
$=80-20=60\,cm$