Question

If position of the particle as a function of time(t) is given
as (x + 2)2 = t (where x is in m, tis in second and t> 0)
then velocity of particle as a function of time will be

Answer 1

Given :

➾ Equation of position of the particle as a function of time has been provided.

$\bigstar\:\underline{\boxed{\bf{\red{(x+2)^2=t}}}}$

To Find :

➳ Velocity of particle as a function of time.

Solution :

$\dashrightarrow\sf\:(x+2)^2=t\\ \\ \dashrightarrow\sf\:x+2=\sqrt{t}\\ \\ \dashrightarrow\bf\:x=\sqrt{t}-2$

✭ Instantaneous velocity of moving object is given by

$\longrightarrow\bf\:v=lim\:{(\Delta t\to 0)\:\dfrac{\Delta x}{\Delta t}=\dfrac{dx}{dt}}\\ \\ \longrightarrow\sf\:v=\dfrac{dx}{dt}=\dfrac{d(\sqrt{t}-2)}{dt}\\ \\ \longrightarrow\sf\:v=\dfrac{1}{2}t^{\frac{1}{2}-1}-0\\ \\ \longrightarrow\underline{\boxed{\bf{\purple{v=\dfrac{1}{2}t^{-\frac{1}{2}}}}}}\:\orange{\bigstar}$

Answer 2

$\bigstar$ $\sf{\red{\underline{\underline{\blue{To\:Find:-}}}}}$

• Velocity(v) of particle as a function of time .

$\bigstar$ $\sf{\blue{\underline{\underline{\red{SOLUTION:-}}}}}$

☞ GIVEN:-

• Position(x) of the particle as a function of time(t) is given as, $\tt{\purple{\boxed{(x\:+\:2)^{2}\:=\:t}}}$

1. x = position [in meter]
2. t = time [in second]
3. “t > 0”

☞ We have know that,

• $\tt{\green{\boxed{{\dfrac{dx}{dt}}\:=\:v}}}$

• [Here, v = Velocity of particle]

☞ Now, we calculate the Derivation of position-time function to get the Velocity of the particle as a function of time .

$\tt{\red{\:(x\:+\:2)^{2}\:=\:t\:}}$

$\tt{\green{\implies\:(x\:+\:2)\:=\:{\sqrt{t}}\:}}$

☞ Now, Derivates the above equation w.r.t time,

$\tt{\blue{\implies\:{\dfrac{d(x)}{dt}}\:+\:{\dfrac{d(2)}{dt}}\:=\:{\dfrac{d({\sqrt{t}})}{dt}}\:}}$

$\tt{\green{\implies\:{\dfrac{dx}{dt}}\:+\:0\:=\:{\dfrac{1}{2{\sqrt{t}}}}\:}}$

$\tt{\purple{\boxed{\implies\:v\:=\:{\dfrac{1}{2}}\:t^{-(\dfrac{1}{2})}\:}}}$

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:v\:=\:{\dfrac{1}{2}}\:t^{-(\dfrac{1}{2})}\:}}}}$