Question

Calculate the acceleration of an object if its velocity is given by v = root 12t-2​

Answer 1

Given :

• v = √(12t) - 2

To Find :

• Value of acceleration (a)

Solution :

We're given that value of velocity is √(12t) - 2 m/s. And we've to find value of acceleration. For this use differentiation

$\implies \sf{a \: = \: \dfrac{dv}{dt}} \\ \\ \implies \sf{a \: = \: \dfrac{d( \sqrt{12}t \: - \: 2)}{dt}} \\ \\ \implies \sf{a \: = \: \dfrac{d(2 \sqrt{3} t \: - \: 2)}{dt}} \\ \\ \implies \sf{a \: = \: 2 \sqrt{3} \dfrac{d}{dt}(t) \: + \: \dfrac{d}{dt} (-2) } \\ \\ \implies \sf{a \: = \: 2 \sqrt{3} \: - \: 0} \\ \\ \implies \sf{a \: = \: 2 \sqrt{3}}$

$\therefore$ Acceleration of the object is $\sf{2 \sqrt{3}}$ m/s²

Answer 2

( Question )

Calculate the acceleration of an object if its velocity is given by v = root 12t-2

$\rule{230}2$

Answer★

$\tt\maltese\: Velocity= \sqrt12t -2\\\tt\maltese acceleration=?$

$\implies\tt \bold a=\frac{dv}{dt}\\\implies\tt a=\frac{d\sqrt12t - 2}{dt}\\\implies\tt a=\frac{d(2\sqrt3t - 2)}{dt}\\\tt\implies a=2\sqrt3\frac{d}{dt}(t)+\frac{d}{dt}-2\\\tt\implies a= 2\sqrt3\\\maltese\large\: Acceleration\:of\:the\:body\:is\:2\sqrt3$