The displacement of particle is given by the relation t-x-4 where x is in meters and t is in seconds- The displacement of particle when its velocity is zero is-

The correct option is C 4 m Given t=√(x 4) x=t^2+4......(i) Then #160; #160; v=dxdt=ddt(x) =ddt(t^2+4) =2t when v=0, ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

1st Equation of Motion

The displacem...

Question

The displacement of particle is given by the relation $t = \sqrt{x - 4}$ where $x$ is in meters and $t$ is in seconds. The displacement of particle when its velocity is zero is:

12 m

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

8 m

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

4 m

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

6 m

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

Open in App

Solution

The correct option is C $4 m$
Given $t = \sqrt{x - 4}$
$x = t^{2} + 4...... (i)$
Then $v = \frac{d x}{d t} = \frac{d}{d t} (x)$
$= \frac{d}{d t} (t^{2} + 4)$
$= 2 t$
when $v = 0, 2 t = 0$
$t = 0$
Putting value of t in (i), we get
$x = 4 m$

Suggest Corrections

Similar questions

Q. The relation

3 t = \sqrt{3 x} + 6

describes the displacement of a particle in one direction where

x

is in meters and

t

in seconds. The displacement when velocity is zero is:

Q. The displacement of a particle moving in a straight line is given by relation

x = 6 + 12 t - 2 t^{2}

. Where 't' is in seconds and 'x' is in meters. The distance covered by particle in first

3 s

The displacement x of a particle moving in one dimension is related to time t by the equation $t = \sqrt{x} + 3$ where x is in metres and t in seconds. The displacement of the particle when its velocity is zero is

Q. The displacement

x

of a particle moving in one dimension under the action of a constant force is related to time

t

by the equation

t = \sqrt{x} + 3

, where

x

is in meter and

t

is in second. Find the displacement of the particle when its velocity is zero.

Q. Displacement x of a particle moving in one dimension is related to time t by the equation

t = \sqrt{x} + 2

. The displacement of the particle when its velocity is zero, is (Here x is in metre and t is in second)

Join BYJU'S Learning Program

The displacement of particle is given by the relation t=√x−4 where x is in meters and t is in seconds. The displacement of particle when its velocity is zero is:

The correct option is C 4 mGiven t=√x−4x=t2+4......(i)Then v=dxdt=ddt(x)=ddt(t2+4)=2twhen v=0,2t=0t=0Putting value of t in (i), we getx=4m

The displacement of particle is given by the relation $t = \sqrt{x - 4}$ where $x$ is in meters and $t$ is in seconds. The displacement of particle when its velocity is zero is:

The correct option is C $4 m$
Given $t = \sqrt{x - 4}$
$x = t^{2} + 4...... (i)$
Then $v = \frac{d x}{d t} = \frac{d}{d t} (x)$
$= \frac{d}{d t} (t^{2} + 4)$
$= 2 t$
when $v = 0, 2 t = 0$
$t = 0$
Putting value of t in (i), we get
$x = 4 m$