Question

if the position (x) of a particle with time 't' as x =2t³-t²+5. what acceleration will we at time t =2 sec.​

Answer 1

Answer:

22 m/s²

Explanation:

dx/dt = (2t³ - t² + 5)/dt

dv = 2(3)(t²) - (2)t+ 0

dv = 6t² - 2t

dv/dt = (6t² - 2t)/dt

da = 6(2)t - 2(1)

da = 12t - 2

At t = 2,

Acceleration = 12(2) - 2 = 22

Answer 2

Answer:

$\underline{\boxed{\large{\red{\bf Acceleration = 22 \: m/s^2}}}}$

Explanation:

Differentiating x = 2t³- t² + 5 w.r.t 't'.

We get,

$\implies \tt \dfrac{dx}{dt} = \dfrac{d}{dt}(2t^3-t^2 +5)$

$\implies \tt \dfrac{dx}{dt} = 2(3)t^{3-1} -t(2) + 5$

$\implies \tt \dfrac{dx}{dt} =6t^2-2t+5$

Velocity (dv) = $\tt \dfrac{dx}{dt}$

$\star \: \boxed{\bf dv = 6t^2- 2t+5}$

$\implies \tt \dfrac{dv}{dt} = 6(2)t - 2 +0$

$\implies \tt \dfrac{dv}{dt} =12t-2$

⇒ Acceleration (da) = $\tt \dfrac{dx}{dt}$

$\star \: \boxed{\bf da = 12t-2}$

At time t = 2 sec,

⇒ Acceleration = 12(2) - 2

⇒ Acceleration = 24 - 2

⇒ Acceleration = 22 m/s²

Hence, the acceleration will be 22m/s².

Additional information :

• When y = xⁿ

$\bf \longrightarrow \dfrac{dy}{dx} = nx^{(n-1)}$