Question

A body, moving in a straight line with an initial velocity of $5{\text{ms}}^{-1}$ and a constant acceleration, covers a distance of $30\text{m}$ in the $3^{rd}$ second. How much distance will it cover in the next $2$ seconds?

• A. $70\text{m}$
• B. $80\text{m}$
• C. $90\text{m}$
• D. $100\text{m}$

Answer 1

The correct option is C $90\text{m}$

Distance covered in $n^{th}$ second is given by
$S_n=u+a(n-\frac{1}{2})$.
Distance covered in $3^{rd}$ second is given by
$30=5+a(3-\frac{1}{2})$,
$\Rightarrow a=10{\text{ms}}^{-2}$.
Distance covered in $4^{th}$ second is given by
$S_4=5+10(4-\frac{1}{2})=40\text{m}$
and $S_5=5+10(5-\frac{1}{2})=50\text{m}$
$\therefore$ Distance covered in next two seconds $=S_4+S_5=40+50=90\text{m}$.
Thus, the correct choice is option c.