(D) Given acceleration $a = \frac{dv}{dt} = 3t^2 + 2t + 2$.To find the velocity $v$,we integrate the acceleration with respect to time: $dv = (3t^2 +

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(D) Given acceleration $a = \frac{dv}{dt} = 3t^2 + 2t + 2$.To find the velocity $v$,we integrate the acceleration with respect to time: $dv = (3t^2 + 2t + 2)dt$.Integrating both sides with the given limits: $\int_{v_0}^{v} dv = \int_{0}^{t} (3t^2 + 2t + 2) dt$.Given $v_0 = 2\,m/s$ at $t = 0$,we have: $v - 2 = [t^3 + t^2 + 2t]_0^2$.Substituting the upper limit $t = 2$: $v - 2 = (2^3 + 2^2 + 2(2)) - 0$.$v - 2 = 8 + 4 + 4 = 16$.$v = 16 + 2 = 18\,m/s$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time in seconds. If the particle starts with an initial velocity $v = 2\,m/s$ at $t = 0$,then the velocity at the end of $t = 2\,s$ is $............m/s$.

A
$12$
B
$14$
C
$16$
D
$18$

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Motion in Straight Line

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Uniformly Accelerated Motion

If the velocity of a particle moving along the $x-$ axis is given by $v = k\sqrt{x}$,then which of the following is true? ($a$ is acceleration)

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The position of a particle moving along the $x$ axis is given by the equation $x = (10 + 6t - 3t^2) \ m$. The distance travelled by the particle from $t = 1 \ s$ to $t = 4 \ s$ is: (in $m$)

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An alpha particle enters a hollow tube of $4 \, m$ length with an initial speed of $1 \, km/s$. It is accelerated in the tube and comes out of it with a speed of $9 \, km/s$. The time for which it remains inside the tube is

Medium

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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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For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

MediumTS EAMCET 2025

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The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time in seconds. If the particle starts with an initial velocity $v = 2\,m/s$ at $t = 0$,then the velocity at the end of $t = 2\,s$ is $............m/s$.

Similar Questions

If the velocity of a particle moving along the $x-$ axis is given by $v = k\sqrt{x}$,then which of the following is true? ($a$ is acceleration)

The position of a particle moving along the $x$ axis is given by the equation $x = (10 + 6t - 3t^2) \ m$. The distance travelled by the particle from $t = 1 \ s$ to $t = 4 \ s$ is: (in $m$)

An alpha particle enters a hollow tube of $4 \, m$ length with an initial speed of $1 \, km/s$. It is accelerated in the tube and comes out of it with a speed of $9 \, km/s$. The time for which it remains inside the tube is

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 $.............\,m$.

For a particle moving along a straight line path, the displacements in the third and fifth seconds of its motion are $10 \ m$ and $18 \ m$ respectively. The speed of the particle at time $t=4 \ s$ is (in $ms^{-1}$)

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