A particle starts moving along a line from ze | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the particle start from point $A$ with initial velocity $u=0$. It accelerates with $f$ for a distance $s_1 = \frac{2}{3}d$ to reach point $B$

Vedclass · Accepted Answer

$\sqrt{3d/f}$

Vedclass · Answer

(C) Let the particle start from point $A$ with initial velocity $u=0$. It accelerates with $f$ for a distance $s_1 = \frac{2}{3}d$ to reach point $B$ with velocity $v_1$. Using the equation $v^2 - u^2 = 2as$,we have: $v_1^2 - 0^2 = 2f(\frac{2}{3}d) \Rightarrow v_1^2 = \frac{4}{3}fd \Rightarrow v_1 = 2\sqrt{\frac{fd}{3}}$. Time taken for the first part $t_1$ is given by $v = u + at$: $v_1 = 0 + ft_1 \Rightarrow t_1 = \frac{v_1}{f} = \frac{2}{f}\sqrt{\frac{fd}{3}} = 2\sqrt{\frac{d}{3f}}$. For the second part of the journey from $B$ to $C$,the distance is $s_2 = \frac{1}{3}d$,initial velocity is $v_1$,and final velocity is $v_2 = 0$. Let the retardation be $a'$. Using $v_2^2 - v_1^2 = 2a's_2$: $0 - \frac{4}{3}fd = 2a'(\frac{1}{3}d) \Rightarrow a' = -2f$. The time taken for the second part $t_2$ is given by $v_2 = v_1 + a't_2$: $0 = v_1 - 2ft_2 \Rightarrow t_2 = \frac{v_1}{2f} = \frac{2\sqrt{fd/3}}{2f} = \sqrt{\frac{d}{3f}}$. The total time $t = t_1 + t_2 = 2\sqrt{\frac{d}{3f}} + \sqrt{\frac{d}{3f}} = 3\sqrt{\frac{d}{3f}} = \sqrt{\frac{9d}{3f}} = \sqrt{\frac{3d}{f}}$.

Similar Questions

$A$ lift is coming down from the $8^{th}$ floor and is just about to reach the $4^{th}$ floor. Taking the ground floor as the origin and the upward direction as positive for all quantities,which one of the following is correct?

The velocity acquired by a body moving with uniform acceleration is $30 \ m/s$ in $2 \ s$ and $60 \ m/s$ in $4 \ s$. The initial velocity is $............. \ m/s$.

$A$ particle moves for $8 \, s$. It first accelerates from rest and then retards to rest. If the retardation is $3 \, \times$ the acceleration, then the time for which it accelerates is: (in $s$)

An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement $(s)-$ velocity $(v)$ graph of this object is

For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module