The motion of a particle along a straight line is described by equation $x = 12 t - t^{3}$ where $x$ is in meter and $t$ is in second. The time instant when velocity of the particle becomes zero is

2 s

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

1 s

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

4 s

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

3 s

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

Open in App

Solution

The correct option is A $2 s$
The position of the particle is given by
$x = 12 t - t^{3}$
$∵ v = \frac{d x}{d t}$
$∴ v = 12 - 3 t^{2}$
$\Rightarrow t^{2} = 4$ (For $v = 0$ )
$\Rightarrow t = 2 s$

Suggest Corrections

Similar questions

The motion of a particle along a straight line is described by equation x=8+12t -t^{3 ,}, where x is in metre and t is in second. What is the retardation of the particle when its velocity becomes zero

Q. The motion of a particle along a straight line is described by equation

x = 8 + 12 t - t^{3}

, where, x is in meter and t is in second. The retardation of the particle when its velocity becomes zero, is

Q. The motion of a particle along a stright line is described by the equation

x = 8 + 12 t - t^{3}

where,

x

is in meter and

t

in second. The retardation of the particle when its velocity beocmes zero, is

Q. the motion of a particle along a straight line is described by equation x=2+12t-t^2,where x is in m and t is in s. The retardation of the particlr when its velocity becomes zero is

Q. The position is a particle along a straight line is denoted by the equation

x = 6 + 12 t - t^{3}

. (Here

x

is in meter and

t

is in second). The magnitude of acceleration of the particle when it is at rest is:-

Join BYJU'S Learning Program

The motion of a particle along a straight line is described by equation x=12t−t3 where x is in meter and t is in second. The time instant when velocity of the particle becomes zero is

The correct option is A 2 sThe position of the particle is given by x=12t−t3 ∵v=dxdt ∴v=12−3t2 ⇒t2=4 (For v=0) ⇒t=2 s

The motion of a particle along a straight line is described by equation $x = 12 t - t^{3}$ where $x$ is in meter and $t$ is in second. The time instant when velocity of the particle becomes zero is

The correct option is A $2 s$
The position of the particle is given by
$x = 12 t - t^{3}$
$∵ v = \frac{d x}{d t}$
$∴ v = 12 - 3 t^{2}$
$\Rightarrow t^{2} = 4$ (For $v = 0$ )
$\Rightarrow t = 2 s$