Question

Find the acceleration of the car
at it = 5 sec

(givan) = v=5t^3+2t+5​

Answer 1

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

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.

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

• 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 ,

$\large {\underline {\bold {\boxed{v = \dfrac{dx}{dt} }}}}$