Question

A body travels a distance of 20 m in the $7^{th}$ second and 24 m in $9^{th}$ second. How much distance shall it travel in the $15^{th}$ second?

• A. 10 m
• B. 16 m
• C. 24 m
• D. 36 m

Answer 1

Answer: (D)

Here, $D_7=20m,D_9=24m,D_{15}=?$
Let $u$ and $a$ be the initial velocity and uniform acceleration of the body.
$D_n=u+\frac{a}{2}(2n-1)$
$\therefore D_7=u+\frac{a}{2}(2\times 7-1)$ or
$20=u+\frac{13a}{2}$
and $D_9=u+\frac{a}{2}(2\times 9-1)$, or
$24=u+\frac{17}{2}a$
Subtracting equation (i) from equation (ii), we get
$4=2a$ or $a=2m{s}^{-2}$
Putting this value in equation (i), we get
$20=u+\frac{13}{2}\times 2=u+13$ or
$u=7m{s}^{-1}$
$\therefore D_{15}=u+\frac{a}{2}(2\times 15-1)$
$=7+\frac{2}{2}\times 29=36m$