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

An electron starting from rest has a velocity that increases linearly with time,given by $v = kt$,where $k = 2 \, m/s^2$. The distance travelled in the first $3 \, s$ will be ........... $m$.

The velocity of a bullet is reduced from $200\; m/s$ to $100\; m/s$ while travelling through a wooden block of thickness $10\; cm$. The retardation,assuming it to be uniform,will be ........... $\times 10^4\; m/s^2$.

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2/3}$. If at an initial point,the velocity is $8 \,m/s$,then the distance travelled by the object before it stops is (in $\,m$)

$A$ lift of mass $M = 500 \, kg$ is descending with a speed of $2 \, ms^{-1}$. Its supporting cable begins to slip,thus allowing it to fall with a constant acceleration of $2 \, ms^{-2}$. The kinetic energy of the lift at the end of a fall through a distance of $6 \, m$ will be $........... \, kJ$.

$A$ body $A$ moves with a uniform acceleration $a$ and zero initial velocity. Another body $B$ starts from the same point and moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module