Question

if displacement of a particle varies with time as
√x = 2t + 7, velocity of particle is proportional to
(1) √x
(2) X
(3) x2
(4) 1/x answer is √x.plss solve fast. urgent​

Answer 1

bro if you wanna know only proportional of velocity then the answer is √x

Answer 2

ANSWER:

• Velocity of particle is proportional to √x.

GIVEN:

• Displacement of a particle varies with time as √x = 2t + 7.

TO FIND:

• The proportionality of velocity.

EXPLANATION:

$\sf \leadsto \sqrt{ x} = 2t + 7$

Squaring on both sides.

$\sf \leadsto x = (2t + 7 {)}^{2}$

$\boxed{ \bold{ \large{ \gray{(A + B)^2=A^2+2AB+B^2}}}}$

$\sf \leadsto x = 4 {t}^{2} + 28t + 49$

$\boxed{ \bold{ \large{\gray{ \frac{d}{dt}x =velocity }}}}$

$\sf \leadsto \dfrac{d}{dt} x = \dfrac{d}{dt}4 {t}^{2} + \dfrac{d}{dt}28t + \dfrac{d}{dt}49$

$\boxed{ \bold{ \large{\gray{ \frac{d}{dt} {x}^{n} =n {x}^{n - 1} }}}}$

$\boxed{ \bold{ \large{\gray{ \frac{d}{dt} k =0}}}}$

$\sf \leadsto v = 8t +28 + 0$

$\sf \leadsto v = 8t +28$

$\sf \leadsto \sqrt{ x} = 2t + 7$

$\sf \leadsto \sqrt{ x} - 7= 2t$

$\sf \leadsto t = \dfrac{ \sqrt{ x} - 7}{2}$

$\sf \leadsto We\ know\ that\ v = 8t +28$

$\sf \leadsto v = 8 \left(\dfrac{ \sqrt{ x} - 7}{2} \right) +28$

$\sf \leadsto v = 4 \left(\sqrt{ x} - 7 \right) +28$

$\sf \leadsto v = 4 \sqrt{ x} - 28 +28$

$\sf \leadsto v = 4 \sqrt{ x}$

$\sf \leadsto v \ \propto \ \sqrt{ x}$

Hence velocity of particle is proportional to √x.