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  • The relation between time and distance for a moving body is given as <img alt="t=m x^{2}+n x" src="https://entrance

The relation between time t and distance x for a moving body is given as t=m x^{2}+n x, where m$ and $\mathrm{n} are constants. The retardation of the motion is : (When v stands for velocity)
 
Option: 1 2 \mathrm{mv}^{3}
Option: 2 2 \mathrm{mnv}^{3}
Option: 3 2 \mathrm{nv}^{3}
Option: 4 2 n^{2} v^{3}

Answers (1)

best_answer

t= mx^{2}+nx
retardation= \frac{-dv}{dt}= \frac{-vdv}{dx}
differentiating throughout w.r.t x
\frac{dt}{dt}= m\, 2x\frac{dx}{dt}+n\frac{dx}{dt}
1= 2mxv+nv\Rightarrow v= \frac{1}{2mx+n}\rightarrow (1)
differentiating throughout w.r.t x
\frac{d\left ( 1 \right )}{x}= 2m\left [x \frac{dv}{dx} +v\frac{dx}{dx}\right ]+n\frac{dv}{dx}
0= 2m\left [x \frac{dv}{dx}+v \right ]+n\frac{dv}{dx}
-\left [ 2mx+n \right ]\frac{dv}{dx}= 2mv
\frac{dv}{dx}= \frac{-2mv}{\left ( 2mx+n \right )}\rightarrow (2)
retardation = \frac{-vdv}{dx}= -v\left ( \frac{-2mv}{2mx+n} \right )
retardation =\frac{2mv^{2}}{\frac{1}{v}}= 2mv^{3}\, \left ( from\, eq\left ( 1 \right ) \right )
The correct option is (1)

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vishal kumar

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