Question

The position of a particle is given by (x = 3t³ + 5t² - 6t) meters. Find the velocity and the acceleration at t = 1 sec.

Answer 1

Explanation:

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Answer 2

Solution:

Given:

=> Position of particle (x) = 3t³ + 5t² - 6t m

=> time = 1 sec

To find:

=> Acceleration

=> Velocity

We know that,

$\sf{Velocity = \dfrac{dx}{dt}=\dfrac{d}{dt}(3t^{3}+5t^{2}-6t)}$

$\sf{\implies \dfrac{d}{dt}(3t^{3})+ \dfrac{d}{dt}(5t^{2})- \dfrac{d}{dt}(6t)}$

$\sf{By\;using\;identity = \dfrac{d}{dx}x^{n}=nx^{n-1}}$

$\sf{\implies \dfrac{d}{dt}(3t^{3})+ \dfrac{d}{dt}(5t^{2})- \dfrac{d}{dt}(6t)}$

$\sf{\implies 3(3t^{3-1})+5(2t^{2-1})-6(t^{1-1})}$

$\sf{\implies 3(3t^{2})+5(2t)-6(t^{0})}$

$\sf{\implies 9t^{2}+10t-6}$

At, t = 1 sec

$\sf{(v)_{t=1}=9(1)^{2}+10(1)-6}$

$\sf{\implies 9 + 10 - 6}$

${\boxed{\boxed{\bf{\implies velocity = 13\;m/s}}}}$

We know that,

$\sf{Acceleration = \dfrac{dv}{dt} = \dfrac{d}{dt}(9t^{2}+10t-6)}$

$\sf{By\;using\;identity = \dfrac{d}{dx}x^{n}=nx^{n-1}}$

$\sf{\implies \dfrac{d}{dt} (9t^{2})+ \dfrac{d}{dt}(10t) - \dfrac{d}{dt}(6)}$

$\sf{\implies 9(2t^{2-1})+10(t^{1-1})-0}$

$\sf{\implies 9(2t)+10}$

$\sf{\implies 18t+10}$

At, t = 1 sec

$\sf{(a)_{t=1}=18(1)+10}$

${\boxed{\boxed{\bf{\implies Acceleration = 28\;m/s^{2}}}}}$