A body starts from the origin and moves along | vedclass

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

Question

(B) Given velocity $v = 4t^3 - 2t$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2$.Position $x = \int v\,dt = \int (4t^3 - 2t)\

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Given velocity $v = 4t^3 - 2t$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2$.Position $x = \int v\,dt = \int (4t^3 - 2t)\,dt = t^4 - t^2$ (assuming $x=0$ at $t=0$).When the particle is at $x = 2\, m$,we have $t^4 - t^2 = 2$,or $t^4 - t^2 - 2 = 0$.Let $u = t^2$,then $u^2 - u - 2 = 0$,which factors as $(u - 2)(u + 1) = 0$.Since $t^2$ must be positive,$t^2 = 2$,so $t = \sqrt{2}\,s$.Substituting $t^2 = 2$ into the acceleration equation: $a = 12(2) - 2 = 24 - 2 = 22\,m/s^2$.

$A$ body starts from the origin and moves along the $X$-axis such that the velocity at any instant is given by $v = (4t^3 - 2t)$,where $t$ is in $s$ and velocity is in $m/s$. What is the acceleration of the particle when it is $2\, m$ from the origin? (in $m/s^2$)

Similar Questions

The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 \ m$ is $0.34 \ ms^{-1}$. If the change in velocity of the body is $0.18 \ ms^{-1}$ during this time,its uniform acceleration is ......... $ms^{-2}$.

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?

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

Draw and explain the $v-t$ graphs for uniformly accelerated motion.

Which of the following represents uniformly accelerated motion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module