Question

it is given that t=px2 + qx where x is displacement and t is time.The acceleration of partical at origin is​

Answer 1

Answer:

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Answer 2

The acceleration of the particle is $\dfrac{-2p}{q^3}$.

Explanation:

We have,

$t=px^2 + qx$......(1)

t is time and x is displacement

Differentiating equation (1) wrt t we get :

$1=\dfrac{d(px^2+qx)}{dt}\\\\1=2px{\cdot} \dfrac{dx}{dt}+q{\cdot} \dfrac{dx}{dt}$

Since, $\dfrac{dx}{dt}=v$ and $\dfrac{dv}{dt}=a$

$1=2pxv+qv$ .......(2)

Now differentiate equation (2) wrt t, we get :

$0=2p\dfrac{d}{dt}(x{\cdot} v)+q\dfrac{dv}{dt}\\\\0=2p(x{\cdot} \dfrac{dv}{dt}+v{\cdot} \dfrac{dx}{dt})+qa\\\\0=2pxa+2pv^2+qa\\\\0=2pv^2+a(2px+q)\\\\a(2px+q)=-2pv^2$

At origin, x = 0, equation (2) becomes, qv = 1

$aq=-2pv^2\\\\a=\dfrac{-2pv^2}{q}\\\\a=\dfrac{-2p(1/q)^2}{q}\\\\a=\dfrac{-2p}{q^3}$

So, the acceleration of the particle is $\dfrac{-2p}{q^3}$.

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