Distance travelled by a particle starting from rest and moving with an acceleration frac{

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2.Motion in Straight Line

easy

The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3}$ $ms^{-2}$ in the third second is

$\frac{{10}}{3}\,m$

$\;\frac{{19}}{3}\,m$

$6\,m$

$4\,m$

(AIPMT-2008)

Solution

$\begin{array}{l}
{\rm{Distance}}\,travelled\,in\,the\,{3^{rd}}\,{\rm{second}}\\
= {\rm{Distance}}\,travelled\,in\,3\,s\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, – \,d{\rm{istance}}\,{\rm{travelled}}\,{\rm{in}}\,\,{\rm{2}}\,\,{\rm{s}}{\rm{.}}\\
As,\,u = 0,\\
{S_{\left( {{3^{rd}}s} \right)}} = \frac{1}{2}a{.3^2} – \frac{1}{2}a \cdot {2^2} = \frac{1}{2}a \cdot 5\\
Give\,a\frac{4}{5}m{s^{ – 2}};\,\,\,\therefore \,\,\,{S_{\left( {{3^{rd}}s} \right)}} = \frac{1}{2} \times \frac{4}{3} \times 5 = \frac{{10}}{3}\,m
\end{array}$

Standard 11

Physics

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1+S_2$ will be made in time

hard

(JEE MAIN-2024)

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medium

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medium

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ sec to that covered in $5\, sec$ is

medium

Speed-time graph of two cars $A$ and $B$ approaching towards each other is shown in figure. Initial distance between them is $60\,m$. The two cars will cross each other after time $……….\,s$

hard

The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3}$ $ms^{-2}$ in the third second is

Solution

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