(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)\

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(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: $a = 12(2) - 2 = 24 - 2 = 22\,m/s^2$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

Difficult

$A$ 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 $m/s^2$)

A
$28$
B
$22$
C
$12$
D
$10$

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Class 11 Questions

Class 11

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Motion in a Straight Line

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Uniformly Accelerated Motion

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)

MediumTS EAMCET 2021

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The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

EasyTS EAMCET 2023

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Derive the equations of uniformly accelerated motion by the graphical method.

Difficult

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Two bodies $A$ and $B$ start from rest from the same point with a uniform acceleration of $2\,m/s^2$. If $B$ starts one second later,then the two bodies are separated,at the end of the next second,by $.............\,m$.

Medium

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$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

Medium

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$A$ 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 velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

Derive the equations of uniformly accelerated motion by the graphical method.

Two bodies $A$ and $B$ start from rest from the same point with a uniform acceleration of $2\,m/s^2$. If $B$ starts one second later,then the two bodies are separated,at the end of the next second,by $.............\,m$.

$A$ car moving along a straight highway with a speed of $126 \; km/h$ is brought to a stop within a distance of $200 \; m$. What is the retardation of the car (assumed uniform in $m/s^2$)?

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