Question

acceleration of a body moving with initial velocity u changes with distance as you is equal to K square root x where K is positive quantity find distance traveled by the body when initial velocity becomes double​

Answer 1

Answer:

The distance traveled by the body when its initial velocity becomes double is $\frac{3u}{2}[\frac{3u}{2k^2}]^{1/3}$

Explanation:

Initial velocity of the body = u

Given, acceleration changes with distance x as

a = k√x, where k is a constant

We know that rate of change of velocity is acceleration

Thus,

dv/dt = a

Also v = dx/dt

thus a can be written as

$a=v\frac{dv}{dx}$

or, $v\frac{dv}{dx}=k\sqrt{x}$

$\implies v{dv}=k\sqrt{x}dx$

$\implies \int_u^{2u} v{dv}=k\int_0^S\sqrt{x}dx$

$\implies \frac{v^2}{2}\Bigr|_u^{2u} =\frac{2}{3} kx^{3/2}\Bigr|_0^S$

$\implies \frac{(2u)^2}{2}-\frac{u^2}{2} =\frac{2}{3} kS^{3/2}$

$\implies \frac{2}{3} kS^{3/2}=\frac{4u^2-u^2}{2}$

$\implies \frac{2}{3} kS^{3/2}=\frac{3u^2}{2}$

$\implies S^{3/2}=\frac{9u^2}{4k}$

$\implies S^{3}=\frac{81u^4}{16k^2}$

$\implies S=[\frac{81u^4}{16k^2}]^{1/3}$

$\implies S=\frac{3u}{2}[\frac{3u}{2k^2}]^{1/3}$

Answer 2

Explanation:

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