Find the acceleration of the carat it = 5 sec(givan) = v=5t^3+2t+5

1.	Find the acceleration of the carat it = 5 sec(givan) = v=5t^3+2t+5
Answer» $\huge {\bold {\underline {\underline{Question : - }}}}$ A car is moving in such a WAY that velocity v changes with TIME t accordingly to the equation , v = 5t³ + 2t + 5 . Then find the acceleration of the car at t = 5 sec. $\huge {\bold{\underline {\underline{<klux>GIVEN</klux> : - }}}}$ Relation between v and t is given by the equation v = 5t³ + 2t + 5 $\huge {\bold {\underline {\underline{To \: find : - }}}}$ Acceleration of the car at t = 5 $\huge {\bold {\underline {\underline{Solution : - }}}}$ From CALCULAS Relation between v and a is given by , $\dag \: \large \underline {\bold {\boxed{ { a = \frac{dv}{dt} }}}}$ Where , a is acceleration v is velocity t is time We have , v = 5t³ + 2t + 5 $: \implies \: a = \dfrac{d(5 {t}^{3} + 2t + 5)}{dt} \\ \\ \ : \implies \: a = \dfrac{d}{dt} (5 {t}^{3} + 2t + 5) \\ \\ : \implies a = \frac{d}{dt} (5 {t}^{3} ) + \dfrac{d}{dt} (2t) + \dfrac{d}{dt} (5) \\ \\ \large \underline {\bold {\boxed{ \dfrac{d}{dt}(constant) = 0}}} \\ \\ : \implies \: a = 5 \dfrac{d}{dt} ( {t}^{3} ) + 2\dfrac{d}{dt} (t) + 0 \\ \\ \large \underline {\bold {\boxed{ \dfrac{d}{dt} (a {}^{n}) = n. {a}^{n - 1} }}} \\ \\ : \implies \: a = 5 \dfrac{d}{dt} (3t {}^{3 - 1} ) + 2\dfrac{d}{dt} (1 {t}^{1 - 1} ) \\ \\ : \implies \: a = 5(3t {}^{2} ) + 2( {t}^{0} ) \\ \\ \large {\bold {\boxed{ {a}^{0} = 1}}} \\ \\ : \implies \: a = 15 {t}^{2} + 2(1) \\ \\ : \implies {\bold {\boxed {\pink {a = 15 {t}^{2} + 2}}}}$ At time t = 5 , $: \implies \: a = 15(5 {}^{2}) + 2 \\ \\ : \implies a = 15(25) + 2 \\ \\ : \implies \: a = 375 + 2 \\ \\ : \implies \: a = 377 \: ms {}^{ - 2}$ ∴ The Acceleration of the car at t = 5 sec is 377 m/s² $\huge {\bold {\underline {\underline{Additional \: info : - }}}}$ ★ Instantaneous power of a BODY is given by , $\large {\underline {\bold {\boxed{P = \frac{dW}{dt} }}}}$ Where , W is work done ★ If we are given the position of a particle at any instant is 'x' , then velocity at any instant is given by , $\large {\underline {\bold {\boxed{v = \dfrac{dx}{dt} }}}}$

Answer»

$\huge {\bold {\underline {\underline{Question : - }}}}$

A car is moving in such a WAY that velocity v changes with TIME t accordingly to the equation , v = 5t³ + 2t + 5 . Then find the acceleration of the car at t = 5 sec.

$\huge {\bold{\underline {\underline{<klux>GIVEN</klux> : - }}}}$

Relation between v and t is given by the equation v = 5t³ + 2t + 5

$\huge {\bold {\underline {\underline{To \: find : - }}}}$

Acceleration of the car at t = 5

$\huge {\bold {\underline {\underline{Solution : - }}}}$

From CALCULAS Relation between v and a is given by ,

$\dag \: \large \underline {\bold {\boxed{ { a = \frac{dv}{dt} }}}}$

Where ,

a is acceleration
v is velocity
t is time

We have ,

v = 5t³ + 2t + 5

$: \implies \: a = \dfrac{d(5 {t}^{3} + 2t + 5)}{dt} \\ \\ \ : \implies \: a = \dfrac{d}{dt} (5 {t}^{3} + 2t + 5) \\ \\ : \implies a = \frac{d}{dt} (5 {t}^{3} ) + \dfrac{d}{dt} (2t) + \dfrac{d}{dt} (5) \\ \\ \large \underline {\bold {\boxed{ \dfrac{d}{dt}(constant) = 0}}} \\ \\ : \implies \: a = 5 \dfrac{d}{dt} ( {t}^{3} ) + 2\dfrac{d}{dt} (t) + 0 \\ \\ \large \underline {\bold {\boxed{ \dfrac{d}{dt} (a {}^{n}) = n. {a}^{n - 1} }}} \\ \\ : \implies \: a = 5 \dfrac{d}{dt} (3t {}^{3 - 1} ) + 2\dfrac{d}{dt} (1 {t}^{1 - 1} ) \\ \\ : \implies \: a = 5(3t {}^{2} ) + 2( {t}^{0} ) \\ \\ \large {\bold {\boxed{ {a}^{0} = 1}}} \\ \\ : \implies \: a = 15 {t}^{2} + 2(1) \\ \\ : \implies {\bold {\boxed {\pink {a = 15 {t}^{2} + 2}}}}$