(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 Straight Line Uniformly Accelerated Motion

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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 $m/s^2$)

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

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Class 11

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Motion in Straight Line

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Uniformly Accelerated Motion

$A$ body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

MediumMHT CET 2025

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$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

Difficult

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$A$ particle moves along a straight line. Its position at any instant is given by $x = 32t - \frac{8t^3}{3}$,where $x$ is in $m$ and $t$ is in $s$. Find the acceleration of the particle at the instant when the particle is at rest $..........\,m/s^2$.

Medium

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In a car race on a straight road,car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to

DifficultJEE MAIN 2019

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$A$ particle moving with a uniform acceleration travels $24 \ m$ and $64 \ m$ in the first two consecutive intervals of $4 \ sec$ each. Its initial velocity is......$m/sec$.

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

$A$ body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the $n^{\text{th}}$ second of its motion to the total distance travelled in $n$ seconds is

$A$ particle starts moving rectilinearly at time $t = 0$ such that its velocity $v$ changes with time $t$ according to the equation $v = t^2 - t$,where $t$ is in seconds and $v$ is in $m/s$. The time interval for which the particle retards is

$A$ particle moves along a straight line. Its position at any instant is given by $x = 32t - \frac{8t^3}{3}$,where $x$ is in $m$ and $t$ is in $s$. Find the acceleration of the particle at the instant when the particle is at rest $..........\,m/s^2$.

In a car race on a straight road,car $A$ takes a time $t$ less than car $B$ at the finish and passes the finishing point with a speed $v$ more than that of car $B$. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then $v$ is equal to

$A$ particle moving with a uniform acceleration travels $24 \ m$ and $64 \ m$ in the first two consecutive intervals of $4 \ sec$ each. Its initial velocity is......$m/sec$.

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