Question

anyone can solve this ?
Question is in attachment​

Answer 1

Relationship between displacement and time is given as :

$x = {t}^{ \frac{1}{2} }$

To find :

Relationship between

• Velocity and Time

• Acceleration and Time

• Acceleration and velocity

Calculation:

$x = {t}^{ \frac{1}{2} }$

$= > v = \dfrac{dx}{dt}$

$= > v = \dfrac{d( {t}^{ \frac{1}{2} }) }{dt}$

$= > v = \dfrac{1}{2} {t}^{ - \frac{1}{2} }$

$= > v = \dfrac{1}{2 \sqrt{t} }$

Again differentiation wrt time :

$= > a = \dfrac{dv}{dt}$

$= > a = \dfrac{1}{2} \dfrac{d( {t}^{ - \frac{1}{2} } )}{dt}$

$= > a = - \dfrac{1}{4} {t}^{ - \frac{3}{2} }$

$= > a = \dfrac{ - 1}{4 {t}^{ \frac{3}{2} } }$

Relating between Acceleration And Velocity by removing term t :

$= > a = \dfrac{ - 1}{4 {t}^{ \frac{3}{2} } }$

$= > a = \dfrac{ - 1}{4 ( { \frac{1}{4 {v}^{2} } )}^{ \frac{3}{2} } }$

$= > a = \dfrac{ - 8 {v}^{3} }{4}$

$= > a = - 2 {v}^{3}$

Answer 2

Answer :

Given :

The relation between displacement and time is given as ::

$\sf \: x = \sqrt{t}$

Explanation :

We have to find,the relation between :

• Velocity and Time

• Acceleration and Time

• Acceleration and Velocity

Differentiating velocity w.r.t time,we get :

$\sf \: v = \dfrac{dx}{dt} \\ \\ \longrightarrow \sf \: v = \dfrac{d( \sqrt{t} )}{dt} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf v = \dfrac{1}{2 \sqrt{t} } }}$

Differentiating acceleration w.r.t time,we get :

$\sf \: a= \dfrac{dv}{dt} \\ \\ \longrightarrow \sf \: a = \dfrac{1}{2} \times \dfrac{d( {t}^{ - \frac{1}{2} } )}{dt} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf a = - \dfrac{1}{4 \sqrt{t^3} } }}$

Squaring v = 1/2√t on both sides,

$\sf v^2 = \dfrac{1}{4t} \\ \\ \longrightarrow \sf t = \dfrac{1}{4v^2}$

Comparing the expressions of Velocity and Acceleration,

$\sf \: a = - \dfrac{1}{2\sqrt{t^2}} \times \bigg( \dfrac{1}{2 \sqrt{t} } \bigg) \\ \\ \longrightarrow \sf a = - \dfrac{v}{2 \times \big(\dfrac{1}{4v^2}\big)}\\ \\ \longrightarrow \: \boxed{ \boxed{ \sf a = - 2v^3}}$