Question

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

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

Answer 1

Answer: (D)

$S_3=u+\frac{a}{2}\left(2\times 3-1\right)=4$ or $u+\frac{5}{2}a=4$
$S_5=u+\frac{a}{2}\left(2\times 5-1\right)=12$ or $u+\frac{9}{2}a=12$
On solving, $u=-6m{s}^{-1},a=4m{s}^{-2}$
Distance travelled in next $3$ seconds $=S_8-S_5$
$=\left[-6\times 8+\frac{1}{2}\times 4\times {\left(8\right)}^2\right]-\left[-6\times 5+\frac{1}{2}\times 4\times {\left(5\right)}^2\right]$
$=80-20=60cm$