Question

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

Answer 1

$\underline{\underline{\Large{\sf{\color{lime}{Given:-}}}}}$

v = 12t - 2

→ Velocity as a function of time ... [v = f(t)]

$\underline{\underline{\Large{\sf{\color{lime}{To\ find:-}}}}}$

The acceleration of the object whose velocity is given by v = 12t - 2.

$\underline{\underline{\Large{\sf{\color{lime}{Solution:-}}}}}$

We can get the acceleration by differentiating the velocity.

$\sf{a=\dfrac{dv}{dt}=\dfrac{d}{dt}f(t) }$

$\sf{\implies f'(t)=\dfrac{d}{dt}(\sqrt{12}t-2) }\\\\\sf{\implies f'(t)=\dfrac{d}{dt}(2\sqrt{3}t-2) }\\\\\displaystyle{\sf{\implies f'(t)= 2\sqrt{3}.\frac{d}{dt}(t)+\frac{d}{dt}(-2) }}\\\\\textsf{(Linear differrentiation)}\\\\\displaystyle{\sf{\implies f'(t)= 2\sqrt{3}.1+0}}\\\\\bullet \sf{Derivative\ of\ the\ differentiation\ variable\ is\ 1.}\\\bullet \sf{Derivative\ of\ a\ constant\ is\ 0.}\\\\\boxed{\displaystyle{\sf{\implies f'(t)= 2\sqrt{3}}}}$

Therefore, the acceleration of the object is 2√3 m/s².

Answer 2

$\red{\underline{\sf{ acceleration \:\:of \:\:the \:\:object\:\: is\:\: 2\sqrt{3} m/s^{2}.}}}$

$\rule{200}2$

$\blue{\huge{\boxed{\sf{SOLUTION:}}}}$

$\boxed{\boxed{\pink{\tt{a=\dfrac{dv}{dt}=\dfrac{d}{dt}f(t)}}}}$

$\green {\tt{\implies f'(t)=\dfrac{d}{dt}(\sqrt{12}t-2) }}$

$\green{\tt{\implies f'(t)=\dfrac{d}{dt}(2\sqrt{3}t-2) }}$

$\green{\displaystyle{\tt{\implies f'(t)= 2\sqrt{3}.\frac{d}{dt}(t)+\frac{d}{dt}(-2) }}}$

$\green{\displaystyle{\sf{\implies f'(t)= 2\sqrt{3}.1+0}}}$

$\bf{Derivative\: of\: the\: differentiation\: variable\: is\: 1.} \\ \bf{Derivative\: of\: a\: constant\: is\: 0.}$

$\red{\huge{\boxed{\displaystyle{\sf{\implies f'(t)= 2\sqrt{3}}}}}}$