A particle starting from rest moves in a straight line under variable acceleration given by $a = 12 t^{2}$ , where $a$ is in $m s^{- 2}$ and $t$ is in second. Distance covered by the particle in the $3$ $s e c o n d s$

65

m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

81

m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

162

m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

256

m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $81$ $m$
$a = 12 t^{2}$
$\frac{d v}{d t} = 12 t^{2}$
$\Rightarrow v = \int 12 t^{2} d t$
$\Rightarrow v = 4 t^{3}$
$\frac{d x}{d t} = 4 t^{3}$
$\Rightarrow x = t^{4}$
Put, $t = 3$
$x = 81$ $m$

Suggest Corrections

Similar questions

Q. A particle moves in a straight line with a velocity

v_{t} = | t - 4 |

m / s

where

t

is time in seconds. The distance covered by the particle in

8 s

Q. A particle starts from rest at

t = 0

and moves in a straight line with acceleration as shown in the following figure.

The velocity of the particle at

t = 3

seconds is

Q. A particle starting from rest undergoes acceleration given by

a = | t - 2 | {m/s}^{2}

where,

t

is time in second. Velocity of particle after

4 sec

Q. The distance travelled by a particle starting from rest and moving with an acceleration

\frac{4}{3} {m/s}^{2}

, in the third second is

A particle starts from rest and moves with an acceleration of $= [4 - | t - 3 |] \frac{m}{s^{2}}$ where $^{'} t^{'}$ is in seconds. The velocity of the particle when it is $2$ m from the origin is

Join BYJU'S Learning Program

A particle starting from rest moves in a straight line under variable acceleration given by a=12t2, where a is in ms−2 and t is in second. Distance covered by the particle in the 3 seconds

The correct option is B 81 ma=12t2dvdt=12t2⇒v=∫12t2dt⇒v=4t3dxdt=4t3⇒x=t4Put, t=3x=81 m

A particle starting from rest moves in a straight line under variable acceleration given by $a = 12 t^{2}$ , where $a$ is in $m s^{- 2}$ and $t$ is in second. Distance covered by the particle in the $3$ $s e c o n d s$

The correct option is B $81$ $m$
$a = 12 t^{2}$
$\frac{d v}{d t} = 12 t^{2}$
$\Rightarrow v = \int 12 t^{2} d t$
$\Rightarrow v = 4 t^{3}$
$\frac{d x}{d t} = 4 t^{3}$
$\Rightarrow x = t^{4}$
Put, $t = 3$
$x = 81$ $m$