Question

physics (motion in a straight line) CLASS 11 pls ans this asap.... If velocity of the particle V=kx^(3/2) , then acceleration verses displacement graph is 1.parabola 2.circle 3.straight line 4.rectangular hyperbola... BEST ANS WILL BE MARKED BRAINLIEST:)

Answer 1

Answer:

• The graph will be Parabola

Given:

1. $\sf v=Kx^{\; 3/2}$

Explanation:

$\rule{300}{1.5}$

From the relation we know,

$\large\bigstar\;{\underline{\boxed{\sf a=\dfrac{dv}{dt}}}}$

Here,

• a Denotes acceleration.
• dv Denotes velocity.
• dt Denotes small interval of time.

Solving,

Multiplying and dividing by dx on the R.H.S.

$\displaystyle\longrightarrow\sf a=\dfrac{dv}{dt}\times \dfrac{dx}{dx}\\\\\\\longrightarrow\sf a=\dfrac{dv}{dx}\times \dfrac{dx}{dt}\\\\\\\longrightarrow\sf a=\dfrac{dv}{dx}\times v \ \ \ \Bigg[\because\Bigg(v=\dfrac{dx}{dt}\Bigg)\Bigg]\\\\\\\longrightarrow\sf a=v\;\dfrac{dv}{dx}\quad\dfrac{\quad}{}[1]$

Substituting the value of velocity.

$\displaystyle\longrightarrow\sf a=Kx^{\;3/2}\;\Bigg[\dfrac{d (Kx^{\;3/2})}{dt}\Bigg]\\\\\\\longrightarrow\sf a=Kx^{\;3/2} \times \Bigg[\dfrac{K\;d(x^{\;3/2})}{dt}\Bigg]\\\\\\\longrightarrow\sf a=Kx^{\;3/2}\times \Bigg[K\; x^{\;(3/2\;-\;1)}\Bigg]\\\\\\\longrightarrow\sf a=Kx^{\;3/2}\times \Bigg[K\; x^{\;(1/2)}\Bigg]\\\\\\\longrightarrow\sf a=\bigg(K\times K\bigg)\Bigg[x^{\;(3/2\;+\; 1/2)}\Bigg]\\\\\\\longrightarrow\sf a=K^{2}\Bigg[x^{\;(4/2)}\Bigg]\\\\\\\longrightarrow\sf a=K^2\;x^2$

Now,

$\displaystyle\longrightarrow\sf a\propto x^2 \\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf a\propto x^2}}}}$

Hence, we can see that acceleration is proportional to the square of displacement, which is same of y = x².

Therefore, the Graph will be Parabolic.

$\rule{300}{1.5}$