A particle with initial velocity v_0 moves with constant acceleration in a straight line. dist

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2.Motion in Straight Line

hard

A particle with initial velocity $v_0$ moves with constant acceleration in a straight line. Find the distance travelled in $n^{th}$ second.

Option A

Option B

Option C

Option D

Solution

For given particle, distance travelled in $n^{\text {th }}$ second is, $d=$

(distance travelled in $n$ seconds) – (distance travelled in $(n-1)$ second)

$=\left(v_{0} n+1 / 2 a n^{2}\right)-\left(v_{0}(n-1)+\frac{1}{2} a(n-1)^{2}\right)\left(\text { From equation } d=v_{0} t+\frac{1}{2} a t^{2}\right)$

$=\left(v_{0} n+\frac{1}{2} a n^{2}\right)-\left(v_{0} n-v_{0}+a / 2\left(n^{2}-2 n+1\right)\right)$

$=\left(v_{0} n+\frac{1}{2} a n^{2}-v_{0} n+v_{0}-\frac{1}{2} a n^{2}+a n-\frac{a}{2}\right)$

$=v_{0}+a_{n}-\frac{a}{2}$

$\therefore d=v_{0}+\frac{a}{2}(2 n-1)$

Standard 11

Physics

The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by

medium

(JEE MAIN-2021)

The displacement of a body in $8\,s$ starting from rest with an acceleration of $20\,cm / s ^2$ is $…………$

easy

A particle is moving in a straight line with initial velocity and uniform acceleration $a$. If the sum of the distance travelled in $t^{\text {th }}$ and $( t +1)^{ th }$ seconds is $100 cm$, then its velocity after $t$ seconds, in $………cm / s$, is