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  • A particle moving with velocity   in the mathematical form - <img alt="\phi(x)=\int_0^{x^2} \sqr

A particle moving with velocity \phi(x)  in the mathematical form -
\phi(x)=\int_0^{x^2} \sqrt{t}  dt   The rate of velocity at  x=\frac{1}{2}  is -
 

Option: 1

2


Option: 2

\frac{2}{3}


Option: 3

\frac{1}{2}


Option: 4

\frac{2}{3}


Answers (1)

best_answer

Mathematical form of velocity is given by from question -
\phi(x)=\int_0^{x^2} \sqrt{t} d t

           =\int_0^{x^2} t^{1 / 2} d t=\left[\frac{t^{3 / 2}}{3 / 2}\right]_0^{x^2}

\phi(x)=\frac{2}{3}\left(x^2\right)^{3 / 2} \\

\phi(x)=\frac{2}{3} x^3
Now,
rate of velocity with respect to x -
\frac{\phi(x)}{d(x)} =\frac{d}{d x}\left(\frac{2}{3} x^3\right)
 =\frac{2}{3} \cdot 3 x^2=2 x^2 \\    
\left.\frac{\phi(x)}{d x}\right|_{\text {at } x=1 / 2} =2\left(\frac{1}{2}\right)^2=2 \times \frac{1}{4} \\
                           = \frac{1}{2}

Posted by

Suraj Bhandari

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