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Question: The relation between time t and distance x is \(t = \alpha x^{2} + \beta ⥂ x,\) where α and β are co...

The relation between time t and distance x is t=αx2+β⥂x,t = \alpha x^{2} + \beta ⥂ x, where α and β are constants. The retardation is (v is the velocity)

A

2αv32\alpha v^{3}

B

2βv32\beta v^{3}

C

2αβv32\alpha\beta v^{3}

D

2β2v32\beta^{2}v^{3}

Answer

2αv32\alpha v^{3}

Explanation

Solution

Differentiating time with respect to distance dtdx=2αx+β\frac{dt}{dx} = 2\alpha x + \beta ⇒ v=dxdt=12αx+βv = \frac{dx}{dt} = \frac{1}{2\alpha x + \beta}

So, acceleration (1) = dvdt=dvdx.dxdt\frac{dv}{dt} = \frac{dv}{dx}.\frac{dx}{dt}

=vdvdx=−v.2α(2αx+β)2=−2α.v.v2=−2αv3v\frac{dv}{dx} = \frac{- v.2\alpha}{(2\alpha x + \beta)^{2}} = - 2\alpha.v.v^{2} = - 2\alpha v^{3}