A body covers $26,28,30,32$ meters in $10^{\text {th }}, 11^{\text {th }}$, $12^{\text {th }}$ and…

A body covers $26,28,30,32$ meters in $10^{\text {th }}, 11^{\text {th }}$, $12^{\text {th }}$ and $13^{\text {th }}$ seconds respectively. The body starts
  1. from rest and moves with uniform velocity
  2. from rest and moves with uniform acceleration
  3. with an initial velocity and moves with uniform acceleration
  4. with an initial velocity and moves with uniform velocity

Solution

The distance covered in $\mathrm{n}^{\text {th }}$ second is $S_{n}=u+\frac{1}{2}(2 n-1) a$
where $u$ is initial velocity $\&$ a is acceleration then
$26=\mathrm{u}+\frac{19 \mathrm{a}}{2}$ $\quad$ $\ldots (i)$
$28=\mathrm{u}+\frac{21 \mathrm{a}}{2}$ $\quad$ $\ldots (ii)$
$30=\mathrm{u}+\frac{23 \mathrm{a}}{2}$ $\quad$ $\ldots (iii)$
$32=\mathrm{u}+\frac{25 \mathrm{a}}{2}$ $\quad$ $\ldots (iv)$
From eqs. (i) and (ii) we get $\mathrm{u}=7 \mathrm{~m} / \mathrm{sec}, \mathrm{a}=2 \mathrm{~m} /$
$\mathrm{sec}^{2}$
$\therefore$ The body starts with initial velocity $\mathrm{u}=7 \mathrm{~m} / \mathrm{sec}$ and moves with uniform acceleration $\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}$ *

Asked in: JEE Mains - Motion In One Dimension - Test 2

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