Question

A body moving in a straight line with constant acceleration of $10{\text{ms}}^{-2}$ covers a distance of $40\text{m}$ in the $4^{th}$ second. How much distance will it cover in the $6^{th}$ second?

• A. $50\text{m}$
• B. $60\text{m}$
• C. $70\text{m}$
• D. $80\text{m}$

Answer 1

The correct option is B $60\text{m}$

Given $a=10m{s}^{-2}$
We know that,
Distance covered by the body in the $n^{th}$ second is given by $S_n=u+\frac{1}{2}a(2n-1)$
According to question
$S_4=u+\frac{1}{2}a(8-1)$
$=u+\frac{7}{2}a$
or $40=u+\frac{7}{2}\times 10$
or $u=40-35$
$=5{\text{ms}}^{-1}$
Distance covered by the body in $6^{th}$ second
$s_6=5+\frac{1}{2}\times 10\times (12-1)$
which gives $s_6=60\text{m}$.
Thus, the correct choice is option (b)