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Q.
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?
Motion in a Straight Line
Solution:
Here, $D_{7}=20 m , D_{9}=24 m , D_{15}=?$
Let $u$ and $a$ be the initial velocity and uniform acceleration of the body.
$D_{n}=u+\frac{a}{2}(2 n-1)$
$\therefore \quad D_{7}=u+\frac{a}{2}(2 \times 7-1)$ or
$20=u+\frac{13 a}{2}$
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=2 a $ or $ a=2 m s ^{-2}$
Putting this value in equation (i), we get
$20=u+\frac{13}{2} \times 2=u+13 $ or
$u=7 m s ^{-1}$
$\therefore D_{15}=u+\frac{a}{2}(2 \times 15-1)$
$=7+\frac{2}{2} \times 29=36 m$