A vehicle starts moving in a straight line wi | vedclass

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

Question

(C) For the first part of the journey: $u = 0, a = 4 \,m/s^2$, time $= t_1$. The velocity attained at the end of time $t_1$ is $v_1 = u + at_1 = 4t_1$

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) For the first part of the journey: $u = 0, a = 4 \,m/s^2$, time $= t_1$. The velocity attained at the end of time $t_1$ is $v_1 = u + at_1 = 4t_1$.The displacement in the first $t_1$ seconds is $s_1 = \frac{1}{2}at_1^2 = \frac{1}{2} 	imes 4 	imes t_1^2 = 2t_1^2$.For the second part of the journey: initial velocity $v_1 = 4t_1$, time $= t_2$, and acceleration $= 0$.The displacement in the next $t_2$ seconds is $s_2 = v_1 t_2 = 4t_1 t_2$.For the third part of the journey: initial velocity $v_1 = 4t_1$, final velocity $= 0$, time interval $= t_1$.The displacement in the third part is $s_3 = \frac{v_1 + v_f}{2} 	imes t_1 = \frac{4t_1 + 0}{2} 	imes t_1 = 2t_1^2$.Total time $T = t_1 + t_2 + t_1 = 2t_1 + t_2 = 10 \,s$, so $t_2 = 10 - 2t_1$.Average velocity $v_{	ext{avg}} = \frac{	ext{Total displacement}}{	ext{Total time}} = \frac{s_1 + s_2 + s_3}{10} = 5.1$.Substituting the values: $5.1 = \frac{2t_1^2 + 4t_1 t_2 + 2t_1^2}{10} = \frac{4t_1^2 + 4t_1(10 - 2t_1)}{10}$.$51 = 4t_1^2 + 40t_1 - 8t_1^2$.$4t_1^2 - 40t_1 + 51 = 0$.Solving the quadratic equation: $t_1 = \frac{40 \pm \sqrt{1600 - 4(4)(51)}}{2(4)} = \frac{40 \pm \sqrt{1600 - 816}}{8} = \frac{40 \pm \sqrt{784}}{8} = \frac{40 \pm 28}{8}$.$t_1 = \frac{68}{8} = 8.5 \,s$ (not possible) or $t_1 = \frac{12}{8} = 1.5 \,s$.Therefore, $t_1 = 1.5 \,s$.

Similar Questions

$A$ particle starting from rest moves with a uniform acceleration and covers $x$ meters in the first $5 \ s$. The same particle will cover the following distance in the next $5 \ s$:

The displacement of a particle after time $t$ is given by $x = (k/b^2)(1 - e^{-bt})$ where $b$ is a constant. What is the acceleration of the particle?

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 ........... $\times 10^4\; m/s^2$.

The same retarding force is applied to stop a train. The train stops after $80 \ m$. If the speed is doubled,then the stopping distance will be

$A$ particle is moving with speed $v = b\sqrt{x}$ along the positive $x$-axis. Calculate the speed of the particle at time $t = \tau$ (assume that the particle is at the origin at $t = 0$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module