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

$A$ bus moving along a straight highway with a speed of $72 \,km/h$ is brought to a halt within $4 \,s$ after applying the brakes. The distance travelled by the bus during this time (assume the retardation is uniform) is . . . . . . $m$.

The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

The distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

$A$ rifle bullet loses $1/20^{th}$ of its velocity in passing through a wooden plank. The least number of planks required to stop the bullet is :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module