Question

Placement of a particle is given by X equal to 20
+ 10t² square calculate the instantaneous velocity t=2s

Answer 1

Answer:

$\boxed{\mathfrak{Instantaneous \ velocity \ (v) = 40 \ m/s}}$

Explanation:

Relation between displacement of particle w.r.t. time is given as:

x = 20 + 10t²

Rate of change of displacement of particle w.r.t. time gives Instantaneous velocity:

$\rm \implies v = \dfrac{dx}{dt} \\ \\ \rm \implies v = \dfrac{d}{dt} (20 + 10 {t}^{2} ) \\ \\ \rm \implies v = 20t$

Instantaneous velocity at t = 2s:

$\rm \implies v = 20 \times 2 \\ \\ \rm \implies v = 40 \: m {s}^{ - 1}$

Answer 2

$\bf{\blue{\underline{\underline{\pink{GIVEN:-}}}}}$

• Displacement of a particle is given by X .

• X = 20 + 10t² .

$\bf{\blue{\underline{\underline{\pink{TO\: FIND:-}}}}}$

• The instantaneous velocity at t = 2s .

$\bf{\blue{\underline{\underline{\pink{SOLUTION:-}}}}}$

We have know that,

• Instantaneous velocity → $\bf\pink{\dfrac{dX}{dt}\:}$

$\rm{\implies\:Velocity_{insta.}\:=\:\dfrac{d(20\:+\:10t^2)}{dt}\:}$

$\rm{\implies\:Velocity_{insta.}\:=\:\dfrac{d(20)}{dt}\:+\:\dfrac{d(10t^2)}{dt}\:}$

$\rm{\implies\:Velocity_{insta.}\:=\:20\times{\dfrac{d(1)}{dt}}\:+\:10\times{\dfrac{d(t^2)}{dt}}\:}$

$\rm{\implies\:Velocity_{insta.}\:=\:20\times{0}\:+\:10\times{2t}\:}$

$\rm{\implies\:Velocity_{insta.}\:=\:0\:+\:10\times{2t}\:}$

$\rm{\implies\:Velocity_{insta.}\:=20\times{t}\:}$

☃️ Now, put the value of “t = 2s”, we get

$\rm{\implies\:Velocity_{insta.}\:=\:20\times{2}\:}$

$\rm\green{\implies\:Velocity_{insta.}\:=\:40\:m/s}$

$\rm\pink{\therefore}$ The instantaneous velocity at t = 2s is “40m/s” .