Question

A body moving with a uniform acceleration crosses a distance of $65m$ in the $5th$ second and $105m$ in $9th$ second. How far will it go in $20s$?

• A. $2040m$
• B. $240m$
• C. $2400m$
• D. $2004m$

Answer 1

The correct option is C $2400m$

Distance covered in $nth$ second $S_n=u+\frac{a}{2}(2n-1)$

where $u$ and $a$ are the initial velocity and acceleration respectively.

Case 1 : $S_n=65$ $n=5$

So, $65=u+\frac{a}{2}(2\times 5-1)$

We get $65=u+4.5a$ .....(1)

Case 2 : $S_n=105$ $n=9$

So, $105=u+\frac{a}{2}(2\times 9-1)$

We get $105=u+8.5a$ .....(2)

Equation (2) -(1), $40=4a$

Or $a=10m/s^2$

So, $65=u+4.5\times 10$ $⟹u=20m/s$

Distance covered in 20 seconds, $S=ut+\frac{1}{2}a{t}^2$

where $t=20s$

$S=20\times 20+\frac{1}{2}\left(10\right)(20{)}^2=2400m$