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 $=S_4+S_5=40+50=90\text{m}$.

Alternate solution:

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
$30=5+a(3-\frac{1}{2})$,
$\Rightarrow a=10{\text{ms}}^{-2}$.
So, the distance covered in next $2\text{sec}$ will be the difference of distance covered by the body in $5\text{sec}$ and distance covered by body in $3\text{sec}$
We will find the difference by using formula $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow (5\times 5+\frac{1}{2}\times 10\times 5^2)-(5\times 3+\frac{1}{2}\times 10\times 3^2)=(150-60)=90\text{m}$